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2 years ago

PLEASE HELP ASAP!!!!!!!!!!

Mathematics

1 answer:

Maurinko [17]2 years ago

6 0

Answer: I think it’s a or d Im stuck sorry I couldn’t help

Step-by-step explanation:

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A house on the market was valued at $295,000. After several years, the value increased by 8%. By how much did the house's value

Harlamova29_29 [7]

Answer: $23,600. ; $318,600

Step-by-step explanation:

From the question, we are informed that a house on the market was valued at $295,000 and after several years, the value increased by 8%.

The value of the increase will be 8% of $295,000. This can be mathematically written as:

= 8% of $295,000

= 8/100 × $295,000

= 0.08 × $295,000

= $23,600

The current value of the house will be:

= $295,000 + $23,600

= $318,600

8 0

3 years ago

Help asap please!!!!

Alex73 [517]

Answer:

D. 84 square feet

Step-by-step explanation:

Hope this helps and have a nice day

5 0

2 years ago

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

2 years ago

What is the axis of symmetry

Verdich [7]

Answer:

its is x=0

Step-by-step explanation:

5 0

3 years ago

How to use Pythagorean theorem to find the right triangle side length

MrRissso [65]

Answer:

You would use a^2 + b^2 = c^2

Step-by-step explanation:

a and b are the shorter side of the triangle whereas c is the hypotenuse or longest side of the triangle

5 0

3 years ago