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KengaRu [80]
3 years ago
7

What would the equation be of the form y=mx​

Mathematics
1 answer:
Ronch [10]3 years ago
4 0

Answer:

y = 1.5x

Step-by-step explanation:

y = mx

slope formula: \frac{y2-y1}{x2-x1}

Points: (x1, y1) = (-2, -3)

(x2, y2) = (4, 6)

To find the equation in y = mx form, we need to find the value of m. m is also known as the slope. To find the value of m, input the given points into the slope formula:

\frac{6-(-3)}{4-(-2)}

Simplify:

6 - (-3) = 6 + 3 = 9

4 - (-2) = 4 + 2 = 6

\frac{9}{6} =\frac{3}{2}=1.5

The value of m is 1.5. Now we can write the equation:

y = 1.5x

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A crater shaped like a cone has a height of 24 yd and a diameter of 72 yd.
mezya [45]
The formula for the volume of a cone is (1/3)*area of the base * height. The base is a circle, whose area is pi times radius^2. Here this is pi * (r^2) = 3.14*(72 / 2 yd)^2 = 4069.44 yd^2. And so the volume is (1/3) * 4069.44 yd^2 * 24yd = 32,555.52 yd^3. I recommend that you find the most common formulas of the geometrical figures and make a flashcard that you can use at any moment.
5 0
3 years ago
A sample of students from an introductory psychology class were polled regarding the number of hours they spent studying for the
Anni [7]

Answer:

8.68,13.16

Step-by-step explanation:

Hint- First we have to calculate the mean and standard deviation of the sample and then applying formula for confidence interval we can get the values.

Mean of the sample is,

\mu=\dfrac{\sum _{i=1}^{24}a_i}{24}=\dfrac{262}{24}=10.92

Standard deviation of the sample is,

\sigma =\sqrt{\dfrac{\sum _{i=1}^{24}\left(x_i-10.92\right)^2}{24-1}}=5.6

The confidence interval will be,

=\mu \pm Z\dfrac{\sigma}{\sqrt{n}}

Here,

Z for 95% confidence interval is 1.96, and n is sample size which is 24.

Putting the values,

=10.92 \pm 1.96\cdot \dfrac{5.6}{\sqrt{24}}

=10.92 \pm 2.24

=8.68,13.16

Confidence interval is used to express the degree of uncertainty associated with a sample.

95% confidence interval means that if we used the same sampling method to select different samples and calculate an interval, we would expect the true population parameter to fall within the interval for 95% of the time.

5 0
3 years ago
Find the solution of the given initial value problem. ty' + 6y = t2 − t + 1, y(1) = 1 6 , t > 0
gtnhenbr [62]

Answer:

Step-by-step explanation:

Given is a differential equation as

ty' + 6y = t^2 - t + 1, y(1) = 1 6 , t > 0

Divide this by t to get in linear form

y'+6y/t = t-1+1/t

This is of the form

y' +p(t) y = Q(t)

where p(t) = 1/te^(\int 1/tdt) = t

So solution would be

yt = \int t^2-t+1 dt\\= t^3/3-t^2/2+t+C

siubstitute y(1) = 16

16 = 16^3/3-128+1+C\\C = -1206

4 0
3 years ago
If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf
jonny [76]
With the curve C parameterized by

C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k

with 0\le t\le\dfrac\pi2, and given the vector field

\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k

the work done by \mathbf f on a particle moving on along C is given by the line integral

\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt

where

\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt

The integral is then

\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt
=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t
=269
6 0
3 years ago
Molly spends 5 hours a week in art class. How many minutes does she spend in art class every week?
vampirchik [111]

Answer:

she spends 300 minutes in art class a week

Step-by-step explanation:

5 x 60 is 300

7 0
3 years ago
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