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4 years ago

11

Which of the following demonstrates the Distributive Property?

2 answers:

AnnyKZ [126]4 years ago

8 0

Answer:

THE answer is d

Step-by-step explanation:

Nata [24]4 years ago

4 0

4(2a+3) = 8a+ 12 This represents the distribution property because it multiplies everything in the parentheses. 4 X 2a = 8a 4X3=12

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What is the formula of x

miv72 [106K]

Answer:

Isolate "x" on one side of the algebraic equation by dividing the number that appears on the same side of the equation as part of "x." For example, in the equation "12x = 24", rewrite the equation as "x = 24 / 12" and solve for "x." The solution is "x = 2."

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

MaryAnne has 90 quilt squares. She plans to use 32 of them to make a quilt. What percent of her quilt square stock is she using?

Gemiola [76]

Percentage = ( number used / total number) x 100

32/90 = 0.3555

0.3555 x 100 = 35.56%

Rounded to nearest tenth = 35.6%

4 0

3 years ago

Samantha mixes some amount of 25% sugar syrup with x grams of 10% sugar syrup.

kondor19780726 [428]

Answer:

40 grams

Step-by-step explanation:

y * 25% + x * 10% = 120 * 15%

x = 120 - y

(120-y) * 10% = x * 10%

y * 25% + (120 - y ) * 10% = 120 * 15%

0.25y + 12 - 0.1y = 18

0.15y = 6

y = 6 / 0.15

y = 40

4 0

2 years ago

Find the equation of the line tangent to the graph of

garik1379 [7]

Answer:

$\displaystyle y=\frac{2\sqrt{3}}{15}x+\frac{\pi-2\sqrt{3}}{6}$

Step-by-step explanation:

We want to find the equation of the line tangent to the graph of:

$\displaystyle y=\sin^{-1}\big(\frac{x}{5}\big)\text{ at } x=\frac{5}{2}$

So, we will find the derivative of our equation first. Applying the chain rule, we acquire that:

$\displaystyle y^\prime=\frac{1}{\sqrt{1-(\frac{x}{5})^2}}\cdot\frac{1}{5}$

Simplify:

$\displaystyle y^\prime=\frac{1}{5\sqrt{1-\frac{x^2}{25}}}$

We can factor out the denominator within the square root:

$\displaystyle y^\prime =\frac{1}{5\sqrt{\frac{1}{25}\big(25-x^2)}}$

Simplify:

$\displaystyle y^\prime=\frac{1}{\sqrt{25-x^2}}$

So, we can find the slope of the tangent line at <em>x</em> = 5/2. By substitution:

$\displaystyle y^\prime=\frac{1}{\sqrt{25-(5/2)^2}}$

Evaluate:

$\displaystyle y^\prime=\frac{1}{\sqrt{75/4}}=\frac{1}{\frac{5\sqrt{3}}{2}}=\frac{2\sqrt{3}}{15}$

We will also need the point at <em>x</em> = 5/2. Using our original equation, we acquire that:

$\displaystyle y=\sin^{-1}(\frac{1}{2})=\frac{\pi}{6}$

So, a point is (5/2, π/6).

Finally, by using the point-slope form, we can write:

$\displaystyle y-\frac{\pi}{6}=\frac{2\sqrt{3}}{15}(x-\frac{5}{2})$

Distribute:

$\displaystyle y-\frac{\pi}{6}=\frac{2\sqrt{3}}{15}x+\frac{-\sqrt{3}}{3}$

Isolate. Hence, our equation is:

$\displaystyle y=\frac{2\sqrt{3}}{15}x+\frac{\pi-2\sqrt{3}}{6}$

7 0

3 years ago

Evaluate the expression for the given values of the variables.

Dahasolnce [82]

4(2m-n)-3(2m-n)

4(2(-15)- -18)-3(2(-15)- -18)
4(-30- -18)-3(-30- -18)
4(-12) -3(-12)
-48- -36
-12

The answer is C

4 0

3 years ago