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2 years ago

Simplify. Cubed root of -128x^6y^5

Mathematics

1 answer:

sweet [91]2 years ago

3 0

Answer

-640x^6y

Step-by-step explanation:

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What is the simplified form of the following expression

Rasek [7]

The second option is the answer

7 0

2 years ago

A person is standing 50 ft from a statue. The person looks up at an angle of elevation of 8 degrees when staring at the top of t

borishaifa [10]

The height of the statue is 19.5 feet.

Why?

We can solve the problem using trigonometric formulas. In this case, we are going to use the trigonometric formula of the tangent.

We know that the person is standing 50ft from the statue, so, it will be the base of the two triangles formed by both angles (elevation and depression)

Using the trignometric formula, we have:

First triangle:

$tg(\alpha )=\frac{h_{1}}{base} \\\\tg(8\°)=\frac{h_{1}}{50ft}\\\\0.14*50ft=h_1\\\\h_1=7ft$

Second triangle:

$tg(\alpha )=\frac{h_{2}}{base} \\\\tg(14\°)=\frac{h_{1}}{50ft}\\\\0.25*50ft=h_1\\\\h_1=12.5ft$

Now, the total height of the statue will be:

$TotalHeight=h_1+h_2=7ft+12.5ft\\\\TotalHeight=19.5ft$

Have a nice day!

8 0

3 years ago

Find all the missing sides and angles of this triangle,

iren2701 [21]

Answer:

See solutions below

Step-by-step explanation:

From the given diagram;

AC = opposite

AB = 7 = hypotenuse

Angle of elevation = 70 degrees

Using SOH CAH TOA

Sin theta = opp/hyp

Sin theta = AC/AB

Sin 70 = AC/7

AC = 7sin70

AC = 7(0.9397)

AC = 6.58

Similarly

tan 70 = AC/BC

tan 70 = 6.58/BC

BC = 6.58/tan70

BC = 6.58/2.7475

BC = 2.39

tan m<A = BC/AC

tanm<A = 2.39/6.58

tan m<A = 0.3632

m<A = 19.96degrees

8 0

2 years ago

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

seropon [69]

Answer:

$h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

Step-by-step explanation:

1) The Fundamental Theorem of Calculus in its first part, shows us a reciprocal relationship between Derivatives and Integration

$g(x)=\int_{a}^{x}f(t)dt \:\:a\leqslant x\leqslant b$

2) In this case, we'll need to find the derivative applying the chain rule. As it follows:

$h(x)=\int_{a}^{x^{2}}\sqrt{5+r^{3}}\therefore h'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left (\int_{a}^{x^{2}}\sqrt{5+r^{3}}\right )\\h'(x)=\sqrt{5+r^{3}}\\Chain\:Rule:\\F'(x)=f'(g(x))*g'(x)\\h'=\sqrt{5+r^{3}}\Rightarrow h'(x)=\frac{1}{2}*(r^{3}+5)^{-\frac{1}{2}}*(3r^{2}+0)\Rightarrow h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

3) To test it, just integrate:

$\int \frac{3r^{2}}{2\sqrt{r^3+5}}dr=\sqrt{r^{3}+5}+C$

5 0

2 years ago

Which figure shows a reflection of pre-image DEFG over the x-axis?

Sergio [31]

The image that shows the reflection of DFEG over the x-axis is the FIRST option. The x-axis is the mirror line. The distance from the mirror line to the reflected shape equals to the distance from the original image to the mirror line.

3 0

3 years ago