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Sphinxa [80]
3 years ago
7

I got to hurry so i kinda didnt have time

Mathematics
1 answer:
k0ka [10]3 years ago
5 0

Answer:

b

Step-by-step explanation:

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Sally is having a problem with her puppy leaving the yard so she decides to build a new fence. The length of the yard is 10 feet
Cerrena [4.2K]

The length of the yard is 25ft.

In order to find this we first need to set up variables for the length and the width. Since we don't know anything about the width, we'll set it as x. Then, we know the length is equal to ten more than three times the width. Since the width is x, we can write the length as 3x + 10. Now we can use the formula for the perimeter of a rectangle to solve for x.

2l + 2w = P

2(3x + 10) + 2(x) = 60

6x + 20 + 2x = 60

8x + 20 = 60

8x = 40

x = 5

Now that we have a value for x, we can plug into the length equation and find the length.

3x + 10

3(5) + 10

15 + 10

25

8 0
3 years ago
Which of the following are roots of the polynomial function? check all that apply F(x)=x^3-x^2-5x-3
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Sjjnan snsnsn Jan’s. Ndndnd
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3 years ago
Hi, I need help with equations :D
lana [24]
3(3) - (-8)/4 = 9 + 2 = 11 so (3,-8) is solution

3(4) - 4/4 = 12 - 1 = 11 so (4,4) is solution

answer
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8 0
3 years ago
For the equation 2(4+x)+(13+x)=3x+k which value of k will create an equation with no solutions. .A. x B. 3x .C 15 D. 21
Artist 52 [7]
<h3>Answer: choice C) 15</h3>

Simplify the left side to get

2(4+x)+(13+x)

2(4)+2(x) +13+x

8+2x+13+x

3x+21

------------

So the original equation

2(4+x)+(13+x) = 3x+k

turns into

3x+21 = 3x+k

------------

Subtract 3x from both sides

3x+21 = 3x+k

3x+21-3x = 3x+k-3x

21 = k

k = 21

-----------

If k = 21, then the original equation will have infinitely many solutions. This is because we will end up with 3x+21 on both sides, leading to 0 = 0 after getting everything to one side. This is a true equation no matter what x happens to be.

If k is some fixed number other than 21, then there will be no solutions. This equation is inconsistent (one side says one thing, the other side says something different). If k = 15, then

3x+21 = 3x+k

3x+21 = 3x+15

21 = 15 .... subtract 3x from both sides

The last equation is false, so there are no solutions here.

note: if you replace k with a variable term, then there will be exactly one solution.

6 0
3 years ago
For what value of k will the lines x+2y=0, 3x-4y-10=0 and 5x+ky-7=0 are concurrent?​
konstantin123 [22]

Answer:

After solve the equations we get value of k=3

Step-by-step explanation:

We need to find value of k for which the lines x+2y=0, 3x-4y-10=0 and 5x+ky-7=0 are concurrent.

If the lines are concurrent, they pass through same point.

Let:

x+2y=0--eq(1)\\ 3x-4y-10=0--eq(2)\\ 5x+ky-7=0--eq(3)

First solving equation 1 and 2 to find values of x and y

From eq(1) we find value of x and put it in eq(2)

From \ eq(1) x+2y=0\\x=-2y\\Put x=-2y \ in \ eq(2)\\3x-4y-10=0\\3(-2y)-4y-10=0 \\-6y-4y=10\\-10y=10\\y=\frac{10}{-10}\\y=-1

After solving we get value of y=-1

Now putting in eq(1) to get value of x

x+2y=0\\x+2(-1)=0\\x-2=0\\x=2

So, Value of x= 2

Now put value of x=2 and y=-1 into eq(3) to find value of k

5x+ky-7=0\\5(2)+k(-1)-7=0\\10-k-7=0\\-k+3=0\\-k=-3\\k=3

So, After solve the equations we get value of k=3

4 0
3 years ago
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