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3 years ago

What is the volume of a cube 1 1/2 inches wide 2 1/2 inches long and 4 inches high?

Mathematics

2 answers:

mel-nik [20]3 years ago

6 0

Answer:

231

Step-by-step explanation:

first of all it must be a cuboid bcoz a cube's edge is same i.e length =breadth= height.

now volume of cuboid =lbh

11/2×21/2×4

=231

ans....

Brrunno [24]3 years ago

5 0

Answer:

15 inches

Step-by-step explanation:

V = lwh

V = $\frac{3}{2}$ · $\frac{5}{2}$ · $\frac{4}{1}$ = $\frac{60}{4}$ = 15

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3 years ago

Help please, area of a triangle

morpeh [17]

Answer:

84 ft^2.

Step-by-step explanation:

Use Pythagoras to find the value of h:

25^2 = h^2 + 7^2

h^2 = (25-7)(25+7)

h^2 = 576

h = √576 = 24

Area = 1/2 * 7 * 24

= 84.

4 0

3 years ago

Find sin(2x), cos(2x), and tan(2x) from the given information.

larisa86 [58]

Since $\cot(x)=\frac{2}{3}$ and $\cot^{2} x+1=\csc^{2} x$ , we know that:

$\left(\frac{2}{3} \right)^{2}+1=\csc^{2} x\\\\\frac{13}{9}=\csc^{2} x\\\\\csc x=\frac{\sqrt{13}}{3}$

If $\csc x=\frac{\sqrt{13}}{3}$ , this means that $\sin x=\frac{3}{\sqrt{13}}$ and by the Pythagorean identity,

$\sin^{2} x+\cos^{2} x=1\\\left(\frac{3}{\sqrt{13}} \right)^{2}+\cos^{2} x=1\\\frac{9}{13}+\cos^{2} x=1\\\cos^{2} x=\frac{4}13}\\\cos x=\frac{2}{\sqrt{13}}$

Using the double angle formula for sine, $\sin(2x)=2\left(\frac{3}{\sqrt{13}} \right)\left(\frac{2}{\sqrt{13}} \right)=\boxed{\frac{12}{13}}$
Using the double angle formula for cosine, $\cos(2x)=1-2\left(\frac{3}{\sqrt{13}} \right)^{2}=\boxed{-\frac{5}{13}}$
So, since tan=sin/cos, $\tan (2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{\frac{12}{13}}{-\frac{5}{13}}=\boxed{-\frac{12}{5}}$