$\displaystyle\\\frac{3d-2}{8} =-d+16\frac{1}{4} \\\\\frac{3d-2}{8} =-d+\frac{16*4+1}{4} \\\\\frac{3d-2}{8} =-d+\frac{64+1}{4} \\\\\frac{3d-2}{8} =-d+\frac{65}{4}$

Multiply both parts of the equation by 8:

$\displaystyle\\3d-2=(-d+\frac{65}{4} )(8)\\\\3d-2=-8d+65*2\\\\3d-2=-8d+130\\\\3d-2+2=-8d+130+2\\\\3d=-8d+132\\\\3d+8d=-8d+132+8d\\\\11d=132\\$

Divide both parts of the equation by 11:

$d=12$

8 0

1 year ago

20 points!

meriva

All circles are similar because all circle have same shape.

However Two circles are congruent , if and only if their radii are congruent.

In our question we are asked when are all circles similar given if their radii are congruent.

So we can say this statement is false because no matter the radii are congruent or not , two or more circles are always similar because of their same shape no matter what the measure of their radii is. Only if we are asked when they are congruent, we will consider the radii part.

Answer is false.

4 0

3 years ago

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5 x 1= 5 5x10=50 5x100=500 5x1000= 5000 What pattern do you notice ?

storchak [24]

Adds a zero each time?

7 0

2 years ago

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