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2 years ago

(-4,2) and (3,-5) the slope of the line, M, is?

Mathematics

1 answer:

lukranit [14]2 years ago

8 0

Answer:

13212

Step-by-step explanation:

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Simply the expression 2(10)+2(x+4)

postnew [5]

Step-by-step explanation:

2(10)+2(x+4)

20 + 2x + 8

20 +8 +2x

28+2x

5 0

2 years ago

Read 2 more answers

You buy a package of 12 pencils for $2.64 write this rate as a unit rate?

iogann1982 [59]

Unit rate means how much does it cost for 1 unit

2.64 for 12 pencils
find 1 pencil
divide both sides by 12
0.22 for 1 pencil

0.22 per pencil

3 0

2 years ago

Write the slope- intercept form of the equation of the line passing through the point (3,5) and parallel to the line y=3x+6.

seropon [69]

Answer: The gradient of the line passing through is 3, thus the equation of the line becomes y=3x-4.

Step-by-step explanation: The parallel lines gradients that have similar gradients, so now we have the gradient of 3 because its similar to the other gradient. To get the equation of the line we use this equation: y=mx+c, we have to substitute the x and y to get the constant c.

5=3(3)+c

c= -4

y=3x+c

3 0

2 years ago

Please help me write this in y=mx+b (show work please)

drek231 [11]

Answer:

y=5/2 +10

Step-by-step explanation:

Find the slope.: Do 15-10 OVER 2-0. This gets you the slope 5/2

Find the y intercept: The y intercept is easy. It is just when the x is 0. You can also think of the y axis as when the line TOUCHES the y axis .

4 0

1 year ago

Find the average value of f(x+y)=sin(x+y) over

Troyanec [42]

For each given region $D$ , the average value of $f$ over $D$ is the integral of $f$ over $D$ divided by the area of $D$ . In both cases, $D$ is a rectangle so the area is trivial.

a. The area of $D$ is $\dfrac\pi3\cdot\dfrac{5\pi}3=\dfrac{5\pi^2}9$ . The integral of $f$ over $D$ is

$\displaystyle\int_0^{5\pi/3}\int_0^{\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$\displaystyle=\int_0^{5\pi/3}\left(\cos y-\cos\left(y+\dfrac\pi3\right)\right)\,\mathrm dy$

$=\sin\dfrac{5\pi}3-\sin\left(\dfrac{5\pi}3+\dfrac\pi3\right) - \sin0+\sin\dfrac\pi3=0$

So the average value of $f$ over this region is 0.

b. The area of $D$ is $\dfrac{2\pi}3\cdot\dfrac{7\pi}6=\dfrac{7\pi^2}9$ . The integral of $f$ is

$\displaystyle\int_0^{7\pi/6}\int_0^{2\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_0^{7\pi/6}\left(\cos y-\cos\left(y+\frac{2\pi}3\right)\right)\,\mathrm dy$

$=\sin\dfrac{7\pi}6-\sin\left(\dfrac{7\pi}6+\dfrac{2\pi}3\right) - \sin0+\sin\dfrac{2\pi}3=\dfrac{\sqrt3}2$

The average vale of $f$ over this region is then $\dfrac{\frac{\sqrt3}2}{\frac{7\pi^2}9}=\dfrac{9\sqrt3}{14\pi^2}$ .

4 0

2 years ago