Write two division equations for 16 times 7/8 = 14

1 answer:

6 0

Two division equations for 16 times 7/8=14 are:
14/(7/8)=16
14/16= 7/8
The reason that these equations work is because the inverse operation of multiplication is division. Any factors of the the multiplication equation can divide the product by the factor and the answer would be the other factor. Example:
AB=C
C/B= A
C/A=B
The product always has to be the first Number of the equation while either factor follows.

You might be interested in

Simplify by factoring. What are the excluded values? Choose all that apply.

makvit [3.9K]

Answer:

x = -2

x = 8

Step-by-step explanation:

Excluded values are the ones which make the denominator zero

3x² + x - 10

3x² + 6x - 5x - 10

3x(x + 2) - 5(x + 2)

(x + 2)(3x - 5)

x² - 6x - 16

x² - 8x + 2x - 16

x(x - 8) + 2(x - 8)

(x - 8)(x + 2)

[(x + 2)(3x - 5)] ÷ [(x - 8)(x + 2)]

(3x - 5)/(x - 8)

So excluded values are 8, -2

6 0

3 years ago

#12 with explanation please

4vir4ik [10]

Answer:

30.01 units

Step-by-step explanation:

Given vertices WXYZ:

W(-4,5), X(2,4), Y(3, -4), Z(-1, -6)

Perimeter is the sum of the side length.

Use distance formula to find each side:

WX = $\sqrt{(2+4)^2+(4-5)^2} =\sqrt{37}$ ≈ 6.08
XY = $\sqrt{(3-2)^2+(-4-4)^2} =\sqrt{65}$ ≈ 8.06
YZ = $\sqrt{(-1-3)^2+(-6+4)^2} =\sqrt{20}$ ≈ 4.47
WZ = $\sqrt{(-1+4)^2+(-6-5)^2} =\sqrt{130}$ $\sqrt{(-1+4)^2+(-6-5)^2}=\sqrt{130}$ ≈ 11.40