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2 years ago

PLSSS HELP 1.Find the surface area 2.Find the volume

Mathematics

2 answers:

antoniya [11.8K]2 years ago

5 0

Answer:

1.106 inches squared 2. 60 inches cubed

Step-by-step explanation:

The surface area is just calculated like this 7.5*2*2 (because there are 2 faces) = 30 then 2*4*2= 16 then 4*7.5*2= 60 then you add them to get 106

For the second all you do is multiply the three values which gets you 60

would be amazing if you picked this as brainliest :)

Nadya [2.5K]2 years ago

3 0

Answer:

Step-by-step explanation:

.Find the surface area

(7,5*4) *2 + (7,5*2) *2 + (4*2) * 2 =

60+ 30+ 16 => 106 inch²

2.Find the volume

7,5 * 2*4 =>60 inch³

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Write an equation for the line that passes through (-1,4) with a slope of -10

Maurinko [17]

Answer:

y = - 10x - 6

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here slope m = - 10, hence

y = - 10x + c ← is the partial equation

To find c substitute (- 1, 4) into the partial equation

4 = 10 + c ⇒ c = 4 - 10 = - 6

y = - 10x - 6 ← equation of line

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2 years ago

For a biology project, you measure the weight in grams and the tail length in millimeters of a group of mice. The correlation is

olga nikolaevna [1]

Answer:

Step-by-step explanation:

Given that for a biology project, you measure the weight in grams and the tail length in millimeters of a group of mice.

The correlation is r = 0.9

If you had measured tail length in centimeters instead of millimeters, what would be the correlation?

For this we must understand the meaning of correlation and formula

Correlation is a measure of linear association between the two variables and always lie between -1 and 1

This is got by dividing covariance of (x,y) by product of std deviation of x and y

So because we changed the units of length correlation coefficient would not change.

It would remain the same as 0.90

8 0

3 years ago

Can someone help me with this

stepan [7]

Answer:

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Step-by-step explanation:

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1 year ago

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

7 0

3 years ago

"Expand the brackets and Simplify"<br>(pls i need help asap)

givi [52]

A) 2(1+2c)

2+4c = 2+4c

B) 6(14r-2t)

= 84r-12t

6 0

2 years ago