2 years ago

Add me on snapchaaaat z3eking ew no I'm lonely man not like that I'm gay ahahahahshs

Mathematics

2 answers:

Harlamova29_29 [7]2 years ago

8 0

Step-by-step explanation:

purr boo tell them and I'm going to follow right know

Aleks04 [339]2 years ago

4 0

Answer:

what

Step-by-step explanation:

what does this mean. I eat bees

You might be interested in

Tell us how the data is

Cloud [144]

What’s the question

4 0

3 years ago

Use the remainder theorem to find the remainder for (x ^5 + 32) divided by (x+2) and state whether or not the binomial is a fact

marta [7]

$x+2$ is a factor of $x^5+32$ because (by the remainder theorem) the remainder upon dividing $x^5+32$ by $x+2$ is $(-2)^5+32=0$ .

There's also the sum of fifth powers formula,

$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$

3 0

3 years ago

Read 2 more answers

What is the volume of the following rectangular prism? 4 units Volume = I units

lord [1]

Answer:

sorry i dont quite understand, were you given more information

Step-by-step explanation:

5 0

3 years ago

Hi, does anyone know the answer to this question? I’m bad at geometry and I’m struggling to answer it.

vodka [1.7K]

The answer is (2) a translation

Please Mark BRAINLIEST!

3 0

3 years ago

1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t

natta225 [31]

The area of the surface given by $\vec r_1(u,v)$ is 1. In terms of a surface integral, we have

$1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

By multiplying each component in $\vec r_1$ by 5, we have

$\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}$

and the same goes for the derivative with respect to $v$ . Then the area of the surface given by $\vec r_2(u,v)$ is

$\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}$

8 0

3 years ago