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3 years ago

I really need help. My teacher said that I could use any number.

Mathematics

1 answer:

katen-ka-za [31]3 years ago

4 0

The keyword is Rise over run
So the 2nd y value minus the first y value and divide it by the 2nd x value minus
(y2-y1)/(x2-x1)
(10-7)/5-3)
the first x value and you should get 3/2
*dont worry the if the number is under and not above it’s not a saying to do to the power of but it’s just saying the order in which to put the x and y values.

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22 less than the quotient of an unknown number and 35 is -25.

Agata [3.3K]

-22+(x/35)=-25
add 22 to both sides
x/35=-3
multiply both sides by 35 to cleaf raction
x=-105

number was -105

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3 years ago

What is the answer 14 - (4 + 3) - 2 =

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Answer:

Step-by-step explanation:

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3 years ago

Which expression is equivalent to 0?<br> (a)(-a)<br> a +(-a)<br> a/-a <br> a - (-a)

cupoosta [38]

Answer:

a - a or (a +(-a))

Step-by-step explanation:

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2 years ago

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The endpoints of are A(2, 2) and B(3, 8). is dilated by a scale factor of 3.5 with the origin as the center of dilation to give

goldfiish [28.3K]

Answer:

The given line segment whose end points are A(2,2) and B(3,8).

Distance AB is given by distance formula , which is

if we have to find distance between two points (a,b) and (p,q) is

= $\sqrt{(p-a)^2+(q-b)^2}$

AB= $\sqrt{(3-2)^2+(8-2)^2}=\sqrt{1+36}=\sqrt{37}$ = 6.08 (approx)

Line segment AB is dilated by a factor of 3.5 to get New line segment CD.

Coordinate of C = (3.5 ×2, 3.5×2)= (7,7)

Coordinate of D = (3.5×3, 3.5×8)=(10.5,28)

CD = AB × 3.5

CD = √37× 3.5

= 6.08 × 3.5

= 21.28 unit(approx)

2. Slope of line joining two points (p,q) and (a,b) is given by

m= $\frac{q-b}{p-a}$

m= $\frac{8-2}{3-2}=6$

As the two lines are coincident , so their slopes are equal.

Slope of line AB=Slope of line CD = 6

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3 years ago

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A sphere of radius r is cut by a plane h units above the equator, where

Anika [276]

Consider the top half of a sphere centered at the origin with radius $r$ , which can be described by the equation

$z=\sqrt{r^2-x^2-y^2}$

and consider a plane

$z=h$

with $0$ . Call the region between the two surfaces $R$ . The volume of $R$ is given by the triple integral

$\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates will help make this computation easier. Set

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

Now, the volume can be computed with the integral

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

You should get

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$

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3 years ago