Rectangle ABCD has a length of 8n – 8 and a width of 3n

25m + 100 - 24m - 75 = 68.

Maksim231197 [3]

25m+100−24m−75=68

Step 1: Simplify both sides of the equation.

25m+100−24m−75=68

25m+100+−24m+−75=68

(25m+−24m)+(100+−75)=68(Combine Like Terms)

m+25=68

Step 2: Subtract 25 from both sides.

m+25−25=68−25

m=43

Answer:

m=43

8 0

3 years ago

1. (SLO) Simplify the expression: 6(b-7)

Sav [38]

1 → B

multiply each term in the parenthesis by the 6 outside

6(b-7) = 6b - 42

2 → D

multiply each term in the parenthesis by the 3 outside

3(4x+6y+2) = 12x+18y+6

3 → A

the value outside the parenthesis is -1, hence each term inside is multiplied by -1

-1(6x+5y) = -6x -5y

3 0

3 years ago

Read 2 more answers

An electrician has a piece of wire that is 4 and 3/8 centimeters long. She divides the wire into pieces that are 1 and 2/3 centi

Vlada [557]

Answer:

She has approximately 3 pieces ( 2.625)

Step-by-step explanation:

To get the number of pieces, we need to divide the length of the wire by the individual length of the pieces.

To get the division done, it would be easier to convert what we have into improper fractions

For the piece of wire, the length which is 4 and 3/8 centimeters long will be 35/8 centimeters

while;

the individual length of each of the piece which is 1 and 2/3 centimeters long will be 5/3 centimeters

The number of pieces is thus 35/8 divided by 5/3 = 35/8 * 3/5 = 21/8 = 2.625 which is approximately 3 pieces

6 0

3 years ago

The equation of a circle is (x + 3)2 + (y + 5)2 = 8. Determine the coordinates of the center of the circle.

Illusion [34]

The equation of the circle is written as: (x - h)² + (y - k)² = r²

(h,k) are the points of the center of the circle.
r is the radius.

In case of the given equation the center is (-3,-5)

Hope this helps :)

8 0

3 years ago

Find the roots of:

Elena-2011 [213]

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{1) \ 2x^3-7x^2+8x-3=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of -3 are 1, -1, 3, +3. So:

| 2 -7 8 -3

1 | 2 -5 3

| 2 -5 3 0

1 | 2 -3

2 -3 0

So the factorization is (x-1)² (2x-3)=0. So:

$\bf{ x_1=x_2=1 \qquad x_2=\dfrac{3}{2} }$

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{2) \ x^3-x^2-4=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of -4 are 1, -1, 2, -2, 4, -4. So:

| 1 -1 0 -4

2 | 2 2

1 2 2 0

So the factorization is (x-2)(x²+x+2)=0 . When calculating the discriminant of the trinomial, it is concluded that it has no roots since the result is negative. So you only have one solution.

$\bf{ 1^2-4(2)(2)=1-16=-15 < 0 \quad \Longrightarrow \quad x=2 }$

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{3) \ 6x^3+7x^2-9x+2=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of 2 are 1, -1, 2, -2. So:

| 6 7 9 2

-2 | -12 10 -2

6 -5 1 0

So the factorization is (x+2)(6x²-5x+1)=0 . The quadratic equation is solved by the general formula:

$\bf{ x_{2, 3}&=\dfrac{5\pm \sqrt{(5)^2-4(6)(1)}}{2(6)}=\dfrac{5\pm \sqrt{25-24}}{12}=\dfrac{5\pm 1}{12} }}$

$\large\displaystyle\text{$\begin{gathered}\sf \begin{matrix} x_1=-2&\ \ \ \ \ \ x_{2}=\dfrac{6}{12} \qquad &\ \ \ x_3=\dfrac{4}{12}\\ &\ \ \ x_2=\dfrac{1}{2} \qquad &x_3=\dfrac{1}{3} \end{matrix} \end{gathered}$}$

6 0

2 years ago

Rectangle ABCD has a length of 8n – 8 and a width of 3n - 1.