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garri49 [273]
3 years ago
9

36d^3+24 solve please.

Mathematics
2 answers:
Oduvanchick [21]3 years ago
5 0
The answer is:

12 (3d^3 + 2)

pls mark as brainlist!!
Sergeeva-Olga [200]3 years ago
3 0

Solution, 36d^3+24:\quad 12\left(\sqrt[3]{3}d+\sqrt[3]{2}\right)\left(3^{\frac{2}{3}}d^2-\sqrt[3]{6}d+2^{\frac{2}{3}}\right)

Steps:

36d^3+24

\mathrm{Factor\:out\:common\:term\:}12,\\12\left(3d^3+2\right)

\mathrm{Factor}\:3d^3+2,\\12\left(\sqrt[3]{3}d+\sqrt[3]{2}\right)\left(\left(\sqrt[3]{3}\right)^2d^2-\sqrt[3]{2}\sqrt[3]{3}d+\left(\sqrt[3]{2}\right)^2\right)

\mathrm{Refine},\\12\left(\sqrt[3]{3}d+\sqrt[3]{2}\right)\left(3^{\frac{2}{3}}d^2-\sqrt[3]{6}d+2^{\frac{2}{3}}\right)

Hope\:This\:Helps!!!

-Austint1414

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Step-by-step explanation:

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A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is .5
Elodia [21]

Answer:

7.3% percentage of the bearings produced will not be acceptable.

Step-by-step explanation:

Consider the provided information.

Average diameter of the bearings it produces is .500 inches. A bearing is acceptable if its diameter is within .004 inches of this target value.

Let X is the normal random variable which represents the diameter of bearing.

Thus, 0.500-0.004<X<0.500+0.004

0.496<X<0.504

The bearings have normally distributed diameters with mean value .499 inches and standard deviation .002 inches.

Use the Z score formula: \frac{X-\mu}{\sigma}

Therefore

\frac{0.496-0.499}{0.002}\leq z\leq \frac{0.504-0.499}{0.002}

\frac{-0.003}{0.002}\leq z\leq \frac{0.005}{0.002}

-1.5\leq z\leq 2.5

Now use the standard normal table and determine the probability of that a ball bearing will be acceptable.

P(-1.5\leq z\leq 2.5)=0.9938-0.0668=0.9270

We need to find the percentage of the bearings produced will not be acceptable.

So subtract it from 1 as shown.

1-0.9270=0.073

Hence, 7.3% percentage of the bearings produced will not be acceptable.

3 0
3 years ago
I need help with 4 and 5
krok68 [10]
4)
area of figure = bh = 8x10 = 80
same as a trapezoid have base 6 and 14 and height 8
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a trapezoid with bases of 6 mm and 14 mm and a height of 8 mm

5)
area of original figure A = bh = 8 x 10 = 80
if double h = 16 and b = 20 then A = 16 x 20 = 320
the area is quadrupled.

answer is C.
The area is quadrupled
3 0
3 years ago
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