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3 years ago

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each month for 6 months laras bank account balance decreased by $84. Witch quantity describes the total change in laras bank acc

ount balance

1 answer:

IgorLugansk [536]3 years ago

5 0

Answer:

Step-by-step explanation:

0

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if Leah makes $9.50 per hour and she works 36 hours this week, how much will her paycheck be before taxes?

Dafna11 [192]

Answer:

Leah makes $342 before tax.

Step-by-step explanation:

9.50 x 36 = 342

Leah makes $342 before tax.

I hope this helped and if it did I would appreciate it if you marked me Brainliest. Thank you and have a nice day!

3 0

3 years ago

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If y=3x+4 were changed to y=5x+4 ,how would the graph of the new function compare with the first one ?

Pepsi [2]

The slope of the line would be positive 5/1 instead of positive 3/1. the y-intercept stays the same.

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3 years ago

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Tickets to a school play cost $3 for each student and $8 for each parent. What would be the best choice for variable(s) when wri

algol13

Best choice for variables.....I would let students be " s " and parents be " p "

7 0

3 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

3 years ago

An angle bisector always creates two acute angles. find a counterexample to show that the conjecture is false.

sweet-ann [11.9K]

An acute angle is an angle that is less than 90°. An angle bisector is a ray drawn along an angle that bisects it into two equal and adjacent parts. Now, if the total angle is, say 270°, which is more than a half circle, it would result to two 135-degree angles. In this case, the angle is no longer acute, but obtuse.

5 0

3 years ago