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Sever21 [200]
3 years ago
5

51.What is the gradient of the graph below?​

Mathematics
1 answer:
krok68 [10]3 years ago
6 0

Answer:

Wher is the graph below

Step-by-step explanation:

how can we answer it correctly?

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What is the most common weight?" 10 points This dot plot shows the weights of backpacks, in kilograms, of 50 sixth-grade student
svetoff [14.1K]

Answer:

4

Step-by-step explanation:

i think its correct

5 0
3 years ago
Logan has a budget of $400 to have family pictures taken. There is a sitting fee of $38. Prints cost $25 per page. How many page
fredd [130]

Answer:

14

Step-by-step explanation:

400-38=362

362/25=14.48

Logan can only print 14 whole pages.

5 0
4 years ago
Read 2 more answers
Suppose f(x,y,z) = x2 + y2 + z2 and W is the solid cylinder with height 7 and base radius 2 that is centered about the z-axis wi
kipiarov [429]

In cylindrical coordinates, <em>W</em> is the set of points

<em>W</em> = {(<em>r</em>, <em>θ</em>, <em>z</em>) : 0 ≤ <em>r</em> ≤ 2 and 0 ≤ <em>θ</em> ≤ 2<em>π</em> and -2 ≤ <em>z</em> ≤ 5}

(A) Then the integral of <em>f(x, y, z)</em> over <em>W</em> is

\displaystyle\iiint_W(x^2+y^2+z^2)\,\mathrm dV = \int_0^{2\pi}\int_0^2\int_{-2}^5r(r^2+z^2)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta

(B)

\displaystyle \int_0^{2\pi}\int_0^2\int_{-2}^5r(r^2+z^2)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta = 2\pi \int_0^2\int_{-2}^5(r^3+rz^2)\,\mathrm dz\,\mathrm dr \\\\\\= 2\pi \int_0^2\left(zr^3+\frac13rz^3\right)\bigg|_{z=-2}^{z=5}\,\mathrm dr \\\\\\= 2\pi \int_0^2\left(\frac{133}3r+7r^3\right)\,\mathrm dr \\\\\\= 2\pi \left(\frac{133}6r^2+\frac74r^4\right)\bigg|_{r=0}^{r=2} \\\\\\= 2\pi \left(\frac{110}3\right) = \boxed{\frac{220\pi}3}

8 0
3 years ago
What two factors multiply to -90 add up to 27
JulsSmile [24]

Answer:

30 and -3

Step-by-step explanation:

30 x -3=-90

30-3=27

7 0
3 years ago
Mr. Mole left his burrow that lies below the ground and started digging his way deeper into the ground,
Vedmedyk [2.9K]

Answer:

A(t) = -1.8t - 4.5

Step-by-step explanation:

Let the starting level of Mr. Mole is 'b'.

Since Mr. Mole is descending with respect to the ground, at the rate = 1.8 meters per minute

After time 't' altitude of Mr. Mole will be = b - 1.8t  

If we represent the altitude of Mr. Mole with A(t).

Then the equation will be,

A(t) = b - 1.8t

Or, A(t) = -1.8t + b

At the time t = 5 minutes, his altitude was 13.5 meters.

So, -13.5 = -1.8×(5) + b

b = -13.5 + 9

b = -4.5

Now we substitute the value of b in the equation.

A(t) = -1.8t - 4.5

3 0
3 years ago
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