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3 years ago

The complement of an angle is 25. What is the measure of an angle

1 answer:

8 0

The complement of an acute angle A in degrees is 90 - A. So if the complement is 25 degrees, the angle would be 90 - 25 = 65 degrees.

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As a fraction in simplest terms, what would you multiply the first number by to get

Ivanshal [37]

1.2 I think that’s the answer because 30 divided by 25 is 1.2

3 0

3 years ago

Find the nonpermissable replacement for the variable in this expression 3x over 5x+2

kotegsom [21]

5x + 2 is a denominator.
A denominator cannot equal zero since division by zero is undefined.
Set the denominator equal to zero and solve for x to find which value of x makes the denominator zero.

5x + 2 = 0

5x = -2

x = -2/5

The non-permissible replacement for the variable is -2/5.

5 0

4 years ago

CAN SOMEONE HELP ME IN THIS INTEGRAL QUESTION PLS

finlep [7]

Due to the symmetry of the paraboloid about the <em>z</em>-axis, you can treat this is a surface of revolution. Consider the curve $y=x^2$ , with $1\le x\le2$ , and revolve it about the <em>y</em>-axis. The area of the resulting surface is then

$\displaystyle2\pi\int_1^2x\sqrt{1+(y')^2}\,\mathrm dx=2\pi\int_1^2x\sqrt{1+4x^2}\,\mathrm dx=\frac{(17^{3/2}-5^{3/2})\pi}6$

But perhaps you'd like the surface integral treatment. Parameterize the surface by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k$

with $1\le u\le2$ and $0\le v\le2\pi$ , where the third component follows from

$z=x^2+y^2=(u\cos v)^2+(u\sin v)^2=u^2$

Take the normal vector to the surface to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial u}=-2u^2\cos v\,\vec\imath-2u^2\sin v\,\vec\jmath+u\,\vec k$

The precise order of the partial derivatives doesn't matter, because we're ultimately interested in the magnitude of the cross product:

$\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|=u\sqrt{1+4u^2}$

Then the area of the surface is

$\displaystyle\int_0^{2\pi}\int_1^2\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\int_0^{2\pi}\int_1^2u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$

which reduces to the integral used in the surface-of-revolution setup.

7 0

3 years ago

vanesa multiplies two integers. When she adds the product to 4, the sum is 0. What can you say about the two integers

Lelu [443]

The Integers Are Positive They Are Not Negative Numbers. If It Was Negative It Would Be -2,-3,-4. While If It's Positive It Should Be 0,1,2,3,4 And So On.

4 0

3 years ago

Find the intercepts. 25 POINTS!!!!!!!!!!!

slamgirl [31]

Answer:

What grade is this??

Step-by-step explanation:

3 0

3 years ago