Kaya's family spends $105 to rent a boat for 7
days. The total cost, c, of the boat rental is
proportional to the number of days, d, the
family rents the boat.
с
e:
A. How much does it cost, in dollars, to
rent the boat for 3 days? SHOW
WORK!
B. Create an equation using "c" and "d" to
represent the proportional relationship

Answer:

4

Step-by-step explanation:

h(k(3))=\(\sqrt{2(3)+7+3}\)

         =4

Step-by-step explanation:

a) The test statistic is (4.4-4)/(1.3/sqrt(70)) = 2.83. The p-value is 0.0023. Since the p-value is less than 0.05, we reject the null hypothesis.

b) The test statistic is (5.3-4.4)/sqrt((1.4^2/106)+(1.3^2/70)) = 4.09. The p-value is less than 0.0001. Since the p-value is less than 0.05, we reject the null hypothesis.

Answer:

\(192x ^{3} y ^{3} \)

According to the probabilities given, it is found that the correct option regarding the independence of the events is given by:

No, P(carry cash) != P(carry cash|have children).

If two events, A and B, are independent, we have that:

\(P(A \cap B) = P(A)P(B)\)

Which also means that:

\(P(A|B) = P(A)\)

\(P(B|A) = P(B)\)

In this problem, we have that:

Since P(carry cash) != P(carry cash|have children), they are not independent.

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There are 120 different ways to order 5 people in line at the ice cream truck.

To find the number of ways to order a group of people in line, we can use the formula for permutations, which is $n!$, where n is the number of people in the group.

In this case, we are ordering 5 people in line at the ice cream truck. Using the formula for permutations, we can calculate the number of ways to order these 5 people as $5! = 120$.

Therefore, there are 120 different ways to order 5 people in line at the ice cream truck.

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Answer:

To convert 2 Bigha into Kattha:

If 1 Bigha = 20 Kattha:

2 Bigha = 2 * 20 Kattha = 40 Kattha

If 1 Bigha = 16 Kattha:

2 Bigha = 2 * 16 Kattha = 32 Kattha

The cost of the cereals and where the family should buy, obtained by simple arithmetic operations are;

A. The cost of 1, 2, 3, and 4 boxes of cereals are;

Corner Store; $4.5, $9, $13.5, $18

Bargain World; $6, $9.5, $13, $16.5

B. When the number of boxes of cereals bought are less than 3, the family shoulds buy from the Bargain World, when the number of cereals bought  are more than 3, the family should buy from the Bargain World Store

Simple arithmetic operations are addition, subtraction, division and multiplication operations.

A. The cost of the boxes of cereal obtained by arithmetic operations are as follows;

The cost of buying 1, 2, 3, and 4 boxes at the Corner store obtained from the graph are;

1 box costs $4.5

2 boxes costs $9

3 boxes costs $13.5

4 boxes costs $18

The cost of buying 1, 2, 3, and 4 boxes at the Bargain World are;

Cost of buying the first box = $6

Cost of buying 2 boxes = $6 + $3.5 = $9.5

Cost of buying 3 boxes = $6 + $3.5 + $3.5 = $13

Cost of buying 4 boxes = $6 + $3.5 + $3.5 + $3.5 = $16.5

B. Based on the costs which is initially higher at the Bargain World store, a family.

The cost of 3 boxes at the Bargain World is $13, while at the Corner Store, it is $13.5

The higher number of boxes are more expensive at the Corner Store than the Bargain Worldstore

Therefore, a family should buy from the Corner Store when they are buying less than 3 boxes of cereal and they should buy from the Bargain World when they are buying more than 3 boxes of cereal

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Answer:

1.3235

Step-by-step explanation:

If you take 2.2 and subtract .8765 from it you get 1.3235. If you take that and subtract it from 2.2 you get .8765.

The dimensions of the area covered by flowers are

Since the garden is a 12-foot-by-9-foot fenced area has walking paths along one of the long sides and one of the short sides with a uniform width x.

Since the garden is a rectangle, its area is A = LW where

So, A = LW

= 12 ft × 9 ft

= 108 ft²

Also, since flowers are planted on the remaining side of the garden with a path of width x on on side, we have that the dimensions of the area covered by flowers is

Since the planted part is rectangle, its area is A' = L'W' where L' 12 - x and W'= 9 - x

So, A' = L'W'

= (12 - x)(9 - x)

Since it is given that he part covered by flowers is half of the entire enclosed area, we have that

A' = A/2

(12 - x)(9 - x) = 108 ft²/2

(12 - x)(9 - x) = 54

Expanding the brackets, we have

108 - 21x + x² = 54

x² - 21x + 108 - 54 = 0

x² - 21x + 54 = 0

Using the quadratic formula, we find x

So, \(x = \frac{-b +/- \sqrt{b^{2} - 4ac} }{2a}\)

where a = 1, b = -21 and c = 54

Substituting the values of the variables into the equation, we have

\(x = \frac{-(-21) +/- \sqrt{(-21)^{2} - 4\times1\times54} }{2\times1}\\= \frac{21 +/- \sqrt{441 - 216} }{2}\\= \frac{21 +/- \sqrt{225} }{2}\\= \frac{21 +/- 15 }{2}\\= \frac{21 + 15 }{2} or x = \frac{21 - 15 }{2}\\= \frac{36}{2} or x = \frac{6}{2}\\x = 18 or x = 3\)

Since x cannot be greater than the greatest dimension, we choose x = 3.

So,

So, the dimensions of the area covered by flowers are

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Answer:

Step-by-step explanation:

Because the units are counting by twos of ounces and fourths of cups you have to find the number in the middle. Since you are using oz we would read the left side's number. The liquid is in-between 4 and 6. If you use your amazing fingers and count. 5 is in-between 4 and 6. So the answer is 5 oz.

The majority of business owners avoid flood insurance.

Inferential statistics enables you to draw conclusions ("inferences") from descriptive statistics, which describes data (such as a chart or graph). Using inferential statistics, you can draw conclusions about a population using data from samples. Using inferential statistics, you can try to discover whether the sample data from a tiny sample of people can forecast whether the medication will be effective for everybody (i.e. the population). There are several techniques to accomplish this, including post-hoc (advanced) testing and generating a z-score, which shows where your data would fall in a normal distribution.

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Answer:

\(\sqrt{89}\)

Step-by-step explanation:

You want to use the distance formula which is \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

So if you just plug in the numbers correctly and solve correctly then you should get \(\sqrt{89}\)

The number of ways are there of choosing ten balls from the box if the number of green balls must be in even numbers =15946 ways

To determine the number of ways of choosing ten balls from the box such that the number of green balls chosen is even, we need to consider the different possible combinations of green balls that could be chosen.

First, we can select 0 green balls, in which case we would need to choose all 10 balls from the 5 red and 8 yellow balls. This can be done in (5+8) choose 10 ways, which is 13 choose 10, or 286 ways.

Alternatively, we could choose 2, 4, or 6 green balls, and then select the remaining balls from the red and yellow balls. To count the number of ways of doing this, we can use the binomial coefficient formula.

The total number of ways of selecting 10 balls from the box is (6+5+8) choose 10, which is 19 choose 10, or 92,378 ways.

Therefore, the number of ways of choosing 10 balls from the box such that the number of green balls chosen is even is the sum of the number of ways of choosing 0, 2, 4, or 6 green balls, which is:

(6 choose 0)(13 choose 10) + (6 choose 2)(5 choose 8)(8 choose 0) + (6 choose 4)(5 choose 6)(8 choose 0) + (6 choose 6)(5 choose 4)*(8 choose 0)

Simplifying this expression gives a total of 15,946 ways.

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Answer: For the first pic, c= 17.20. For the second pic, x= 50.

Step-by-step explanation:

We need to use the Pythagorean theorem for these problems. A^2 + B^2= C^2

For the square (first picture).

It doesn't matter what values we put for A and B, as long as the hypotenuse (denoted in the picture as c) is C.

Let A= 10, and B=14

Plug the values into the equation!

\((10)^2 + (14)^2 = C^2\)

Simplify: \(100 + 196 = C^2\)

\(296 = C^2\)

\(\sqrt{296} = \sqrt{C^2}\)

\(\sqrt{296} = C\)

17.20= C

Therefore, c= 17.20

PICTURE #2:

Complete the problem exactly how you did the first one.

Let A=48, and B=14

Plug those values into the pythagorean theorem...

\((48)^2 + (14)^2 = C^2\)

Simplify: \(2304 + 196 = C^2\)

2500 = C^2

\(\sqrt{2500} =\sqrt{C^2}\)

\(\sqrt{2500} = C\)

50= C

SO, 50 = x

Hope this all helps!!

The solution of the system of equations f(x) = 2^x + 1 and g(x) = 3^x is x = 1

From the question, we have the following parameters that can be used in our computation:

f(x) = 2^x + 1

g(x) = 3^x

The above expression is a system of exponential equations

We are required to solve by graph

That implies that we graph the equations in the system on the same plane and write out ordered pairs from the point of intersection of the equations in the system

Next, we plot the graph

See attachment for the graph of the equations

The ordered pairs of the intersection point is (1, 3)

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I'm sorry if this is incorrect, but I think the answer is \(-2, \sqrt{5}, 4, 3^2\)

Explanation: A negative number will always be the least. 3 squared is actually 3*3 and that would be 9. The square root of 5 is 2.2360679775 which is less than 4 so that would go in the second place. I hope this makes sense! :)

Answer:

ask a Tutor its easier when they tell u the answer

Step-by-step explanation:

Area of the triangle with two sides equal 4 cm and 5 cm and the included angle equal to 30° is

\( = \frac{1}{2} \times 4 \times 5 \times sin30 \\ = \frac{1}{2} \times 4 \times 5 \times \frac{1}{2} \\ = 5 \: cm^{2} \)

Hope it will help :)❤

Answer:

They are wrong

Step-by-step explanation:

They first turned the mixed number into an improper fraction, Then they divided but they ignored the negatives which caused them to get the wrong answer.

The right answer is \(4\frac{4}{5}\)

\(-3\frac{1}{5}\)÷\(-\frac{2}{3}\)

\(-\frac{16}{5}\)÷\(-\frac{2}{3}\)

\(\frac{16}{5}\)÷\(\frac{2}{3}\)

\(\frac{16}{5}\)•\(\frac{3}{2}\)

\(\frac{8}{5}\)•\(3\)

\(\frac{24}{5}\)

\(4\frac{4}{5}\)

The experimental probability of not spinning a 5 from the given bar chart above would be = 0.19

To calculate the probability, the formula that should be used would be given below as follows:

Probability = possible outcome/sample space

where:

possible outcome= 19

Sample space= 20+18+22+21+19 = 100

The probability= 19/100 = 0.19

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Problem 1

There are 23 women that drive out of 12+3+23+8+5+32 = 83 people total

Divide the two values: 23/83 = 0.28 approximately

This says about 28% of the total people are women who drive a car.

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Problem 2

The "given" is a very important keyword in probability. In this case, we're given that the employee takes the bus. That means we soly focus on the "bus" column only. Everything else is irrelevant. This is because we know 100% that whoever we selected takes the bus, and we've shrunk down the population just these employees.

We have 8 males who take the bus out of 12+8 = 20 people who take the bus. Therefore, 8/20 = 0.40 represents the relative frequency of males who take the bus out of all bus commuters. In other words, 40% of the bus riders are male.

Answer:

x=-7

Step-by-step explanation:

add 2x to -x to get 1=x+8, then subtract 8 to get x=-7

Answer:

x = -7

Step-by-step explanation:

1 Regroup terms.

-2x+1=8-x−2x+1=8−x

2 Add 2x2x to both sides.

1=8-x+2x1=8−x+2x

3 Simplify  8-x+2x8−x+2x  to  8+x8+x.

1=8+x1=8+x

4 Subtract 88 from both sides.

1-8=x1−8=x

5 Simplify  1-81−8  to  -7−7.

-7=x−7=x

6 Switch sides.

-7 = x

x=−7

Hope this helps!

-Jerc

Answer:35 you are welcome

Step-by-step explanation:

ok

MPG = 66.8

3  is the answer

you're welcome

Step-by-step explanation:

4 is the answer since the question asks if the student would prefer so to get a better opinion you'd have to ask ever student in all grade

Approximately $334,728.87 should be set aside each year at the end of the period to accumulate $80,000 after 15 years with an interest rate of 10 percent.

To determine the amount that should be set aside each year to accumulate $80,000 after 15 years with an interest rate of 10 percent, we can use the Present Value-Future Value (PV-FV) tables.

First, let's find the Present Value Factor (PVF) for the given interest rate and time period. The PVF represents the value of $1 received or paid in the future, expressed in today's dollars.

From the PV-FV table for an interest rate of 10 percent and a time period of 15 years, we find the PVF to be 0.239.

Next, we divide the desired future value ($80,000) by the PVF to calculate the amount that should be set aside each year:

Amount to set aside each year = Future Value / PVF

= $80,000 / 0.239

≈ $334,728.87

Therefore, approximately $334,728.87 should be set aside each year at the end of the period to accumulate $80,000 after 15 years with an interest rate of 10 percent.

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Answer:

6:7

Step-by-step explanation:

Divide 12 and 14 by 2 to simplify the ratio.

Answer: 6:7

Step-by-step explanation:

The number of red balloons in 12 and the number of blue ones is 14, which can be written as 12:14. 12:14 can be simplified to 6:7

x=-4 if you are in a hurry

The solution of the equation are; x = -16 or x = -4. so option A is correct.

A quadratic equation is the second-order degree algebraic expression in a variable. the standard form of this expression is  ax² + bx + c = 0 where a. b are coefficients and x is the variable and c is a constant.

Given equation;

x²+ 20 x + 100 = 36.

x²+ 20 x + 100 - 36 = 0

x²+ 20 x + 64 = 0.

x²+ (16 + 4)x + 64 = 0.

x²+ 16x + 4x + 64 = 0.

x(x + 16) +4 (x + 16) = 0.

(x+ 4) (x + 16) = 0

Thus, the solution of the equation are; x = -16 or x = -4.

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Answer:

\(f(x)=16x^4+96x^{3}+216x^{2}+216x+81\)

Step-by-step explanation:

The function f(x) = (2x + 3)⁴ is a fourth-degree polynomial function.

It can be expanded using the binomial theorem.

\(\boxed{\begin{minipage}{5cm} \underline{Binomial Theorem}\\\\$\displaystyle (a+b)^n=\sum^{n}_{k=0}\binom{n}{k} a^{n-k}b^{k}$\\\\\\where \displaystyle \binom{n}{k} = \frac{n!}{k!(n-k)!}\\\end{minipage}}\)

Comparing the given function with (a + b)ⁿ:

Substitute these values into the binomial theorem formula:

\(\displaystyle (2x+3)^4=\binom{4}{0}(2x)^{4-0}3^{0}+\binom{4}{1}(2x)^{4-1}3^{1}+\binom{4}{2}(2x)^{4-2}3^{2}+\binom{4}{3}(2x)^{4-3}3^{3}+\\\\\\\phantom{wwww}\binom{4}{4}(2x)^{4-4}3^{4}\)

Solve:

\(\begin{aligned}\displaystyle (2x+3)^4&=\binom{4}{0}(2x)^4\cdot3^0+\binom{4}{1}(2x)^{3}\cdot3^1+\binom{4}{2}(2x)^2\cdot3^2+\binom{4}{3}(2x)^{1}\cdot3^3+\binom{4}{4}(2x)^0\cdot3^4\\\\&=\binom{4}{0}16x^4\cdot1+\binom{4}{1}8x^3\cdot3+\binom{4}{2}4x^2\cdot9+\binom{4}{3}2x\cdot27+\binom{4}{4}\cdot81\\\\&=\binom{4}{0}16x^4+\binom{4}{1}24x^3+\binom{4}{2}36x^2+\binom{4}{3}54x+\binom{4}{4}81\\\\&=1\cdot16x^4+4\cdot24x^3+6\cdot36x^2+4\cdot54x+1\cdot81\\\\&=16x^4+96x^3+216x^2+216x+81\end{aligned}\)

Therefore, the expanded function is:

\(f(x)=16x^4+96x^{3}+216x^{2}+216x+81\)

\( \Large{\boxed{\sf F(x) = (2x + 3)^4 = 16x^4 + 96x^3 + 216x^2 + 216x + 81 }} \)

\( \\ \)

To expand the given function, we will apply the binomial theorem, which is the following:

\(\sf(a+b)^n =\sf\sum\limits_{k=0}^{n} \binom{n}{k}a^{n-k}b^{k} \\ \\ \sf \:Where\text{:} \\ \star \: \sf n \: is \: a \: positive \: integer. \: ( n \in \mathbb{N}) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \star \: k \: is \: a \: positive \: integer \: less \: than \: or \: equal \: to \: n. \: (k \leqslant n, \: k \: \in \: \mathbb{ N}) \\ \\ \\ \sf \star \: \displaystyle\binom{ \sf \: n}{ \sf \: k} \: \sf is \: a \: \underline{binomial \: coefficient} \: and \: is \: calculated \: as \: follows\text{:} \\ \\ \\ \sf \displaystyle\binom{ \sf \: n}{ \sf \: k} = \sf \dfrac{n! }{(n - k)! k ! }\)\( \\ \\\)

\( \\ \)

\( \\ \)

\( \sf F(x) = (\underbrace{\sf 2x}_{\sf a} + \underbrace{3}_{\sf b})^{\overbrace{\sf 4}^{n}} \\ \\ \implies \sf a = 2x \: \: ,b = 3 \: \: ,n = 4 \)

\( \\ \)

\( \\ \)

\( \sf (2x + 3)^4 = \displaystyle\sum\limits_{ \sf k=0}^{ \sf 4} \binom{ \sf 4}{ \sf k}( \sf 2x)^{4-k}(3)^{k} \\ \\ \\ \sf = \binom{ \sf 4}{ \sf 0}( \sf 2x)^{4-0}(3)^{0} + \binom{ \sf 4}{ \sf 1}( \sf 2x)^{4-1}(3)^{1} + \binom{ \sf 4}{ \sf 2}( \sf 2x)^{4-2}(3)^{2} + \binom{ \sf 4}{ \sf 3}( \sf 2x)^{4-3}(3)^{3} + \binom{ \sf 4}{ \sf 4}( \sf 2x)^{4-4}(3)^{4} \\ \\ \\ \sf = \binom{ \sf 4}{ \sf 0}( \sf 2x)^{4}(3)^{0} + \binom{ \sf 4}{ \sf 1}( \sf 2x)^{3}(3)^{1} + \binom{ \sf 4}{ \sf 2}( \sf 2x)^{2}(3)^{2} + \binom{ \sf 4}{ \sf 3}( \sf 2x)^{1}(3)^{3} + \binom{ \sf 4}{ \sf 4}( \sf 2x)^{0}(3)^{4} \\ \\ \\ \sf = \binom{ \sf 4}{ \sf 0}( \sf 16 {x}^{4} ) + \binom{ \sf 4}{ \sf 1}( \sf 24 {x}^{3}) + \binom{ \sf 4}{ \sf 2}( \sf 36x^{2}) + \binom{ \sf 4}{ \sf 3}( \sf 54x) + \binom{ \sf 4}{ \sf 4}(81) \)

\( \\ \)

\( \\ \)

\( \\ \star \: \displaystyle\binom{ \sf 4}{\sf \: 0} = \sf \dfrac{4! }{(4-0)!0 ! } = \dfrac{4!}{4!0!} = \dfrac{4!}{4!} = \boxed{\sf 1} \\ \\ \star\:\displaystyle\binom{ \sf 4 }{ \sf \: 1} =\sf \dfrac{4! }{(4 - 1)!1 ! } = \dfrac{4!}{3!1!}= \dfrac{2\times 3 \times 4}{2 \times 3} = \boxed{\sf 4} \\ \\ \star \: \displaystyle\binom{ \sf 4 }{ \sf \: 2} =\sf \dfrac{4! }{(4-2)!2!} = \dfrac{4!}{2!2!}=\dfrac{2 \times 3 \times 4}{2 \times 2} = \boxed{\sf 6} \\ \\ \star \:\displaystyle\binom{ \sf 4 }{ \sf \: 3}= \sf \dfrac{4! }{(4 - 3)!3!} =\dfrac{4!}{1!3!} = \dfrac{2 \times 3 \times 4}{2 \times 3 } = \boxed{\sf 4}\\ \\ \star \:\displaystyle\binom{ \sf 4 }{ \sf \: 4}= \sf \dfrac{4! }{(4 - 4)!4 !} =\dfrac{4!}{0!4!} = \dfrac{2 \times 3 \times 4}{2 \times 3 \times 4 } = \boxed{\sf 1} \)

\( \\ \)

\( \\ \)

\( \sf (2x + 3)^4 = \binom{ \sf 4}{ \sf 0}( \sf 16 {x}^{4} ) + \binom{ \sf 4}{ \sf 1}( \sf 24 {x}^{3}) + \binom{ \sf 4}{ \sf 2}( \sf 36x^{2}) + \binom{ \sf 4}{ \sf 3}( \sf 54x) + \binom{ \sf 4}{ \sf 4}(81) \\ \\ \\ \sf = (1)(16x^4) + (4)(24x^3) + (6)(36x^2) + (4)(54x) + (1)(81) \\ \\ \\ \boxed{\boxed{\sf = 16x^4 + 96x^3 + 216x^2 + 216x + 81}} \)

\( \\ \\ \\ \)

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To determine whether the mean weight of two distinct turtle species is equal. It would take too much time and money to weigh each turtle because there are thousands of them in each group.

A two-sample t-test is used to determine whether or not genuine population means are equal.

The accompanying null hypothesis is usually used in a two-sample t-test:

\(H_0\): The two population averages are equivalent when 1 = 2.

The alternative hypothesis might have two tails, a left or right tail, or none at all:

Two-tailed \(H_1\): μ1 ≠ μ2 Neither of the two population means is equal.

(Left-tailed) \(H_1\): Population 1 mean is lower than population 2 mean by a factor of 1.

\(H_1\) (right-tailed): Populations 1 and 2 have mean values that are higher than each other.

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Answer:

\(x=7\)

Step-by-step explanation:

\(x =\sqrt{(x+18)}+2\)

subtract 2 from both sides:

\(x -2=\sqrt{(x+18)}\)

square both sides:

\((x -2)^2=x+18\)

expand brackets:

\(x^2-4x+4=x+18\)

subtract x from both sides:

\(x^2-5x+4=18\)

subtract 18 from both sides:

\(x^2-5x-14=0\)

factor:

\(x^2+2x-7x-14=0\)

\(x(x+2)-7(x+2)=0\)

\((x+2)(x-7)=0\)

solve for x:

\(x+2=0\implies x=-2\)

\(x-7=0\implies x=7\)

Now we have found the values of x, input them into the original equation to verify:

when \(x = -2\):

\(\sqrt{(-2 +18)}+2=6\\\\ 6\neq 2\implies \textsf {incorrect}\)

when \(x = 7\):

\(\sqrt{(7+18)}+2=7\\\\ 7=7\implies \textsf {correct}\)

Therefore, the only correct solution is \(x=7\)

Let's find x

\(\\ \rm\rightarrowtail x=\sqrt{x+18}+2\)

\(\\ \rm\rightarrowtail x-2=\sqrt{x+18}\)

\(\\ \rm\rightarrowtail x^2-4x+4={x+18}\)

\(\\ \rm\rightarrowtail x^2-5x-14=0\)

\(\\ \rm\rightarrowtail x^2+2x-7x-14=0\)

\(\\ \rm\rightarrowtail (x+2)(x-7)=0\)