Solve the following equation. Then placed a correct number in the box provided. -13

The length of a football field is 8 yards more than twice the width. If the perimeter of the football field is 166 yards, find t

Kamila [148]

Answer:

Width = 25

Length = 58

Step-by-step explanation:

Let the width = w

Length = 2w + 8

2(w + 2w + 8) = 166 Combine terms in the brackets

2(3w + 8) = 166 Divide both sides by 2

(3w + 8) = 83 Subtract 8 from both sides

3w = 83 - 8

3w = 75 Divide by 3

w = 25

L = 2*25 + 8 Find the Length

L = 50 + 8 = 58

6 0

3 years ago

The volume of a cube is 64 cm3.

Rainbow [258]

Answer:

If the volume of a cube is 64 cm3, then the side length is 4 cm, because 4x4x4 = 64.

If the side length is 4cm, then the area of one side is 16cm2, and the surface area is 96cm2

Let me know if this helps!

4 0

3 years ago

The quotient of a number and two is increased by seven. The result is eleven. What is the number?

MatroZZZ [7]

Answer:

C

Step-by-step explanation:

8 is the correct answer.

8 0

3 years ago

Read 2 more answers

8. The equation of a line is y = 2x = 3. Write the equation of a line<br> parallel to this line.

laila [671]

Answer:

(3/2,3)

Step-by-step explanation:

1.Rewrite the equation

y=3 2x=3

2.Solve the equation

y=3 x=3/2

(x,y)=(3/2,3)

5 0

3 years ago

Please help solve these proofs asap!!!

timama [110]

Answer:

Proofs contained within the explanation.

Step-by-step explanation:

These induction proofs will consist of a base case, assumption of the equation holding for a certain unknown natural number, and then proving it is true for the next natural number.

a)

Proof

Base case:

We want to shown the given equation is true for n=1:

The first term on left is 2 so when n=1 the sum of the left is 2.

Now what do we get for the right when n=1:

$\frac{1}{2}(1)(3(1)+1)$

$\frac{1}{2}(3+1)$

$\frac{1}{2}(4)$

$2$

So the equation holds for n=1 since this leads to the true equation 2=2:

We are going to assume the following equation holds for some integer k greater than or equal to 1:

$2+5+8+\cdots+(3k-1)=\frac{1}{2}k(3k+1)$

Given this assumption we want to show the following:

$2+5+8+\cdots+(3(k+1)-1)=\frac{1}{2}(k+1)(3(k+1)+1)$

Let's start with the left hand side:

$2+5+8+\cdots+(3(k+1)-1)$

$2+5+8+\cdots+(3k-1)+(3(k+1)-1)$

The first k terms we know have a sum of .5k(3k+1) by our assumption.

$\frac{1}{2}k(3k+1)+(3(k+1)-1)$

Distribute for the second term:

$\frac{1}{2}k(3k+1)+(3k+3-1)$

Combine terms in second term:

$\frac{1}{2}k(3k+1)+(3k+2)$

Factor out a half from both terms:

$\frac{1}{2}[k(3k+1)+2(3k+2]$

Distribute for both first and second term in the [ ].

$\frac{1}{2}[3k^2+k+6k+4]$

Combine like terms in the [ ].

$\frac{1}{2}[3k^2+7k+4$

The thing inside the [ ] is called a quadratic expression. It has a coefficient of 3 so we need to find two numbers that multiply to be ac (3*4) and add up to be b (7).

Those numbers would be 3 and 4 since

3(4)=12 and 3+4=7.

So we are going to factor by grouping now after substituting 7k for 3k+4k:

$\frac{1}{2}[3k^2+3k+4k+4]$