32.determine the rate of change of the graph

Mathematics

2 answers:

Anon25 [30]3 years ago

7 0

Rise over run 4/2 or 2 would be the rate of change the equation would be
Y=2x+12 or Y=4/2x+12 either one would work

densk [106]3 years ago

6 0

12, Y=2 X=1, starts at 12, rate is (12+(2*Y) + (1*X) )

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Multiply. <br> -3.9(-5.7)(3) <br> The product is:

scZoUnD [109]

Answer:

66.69

Step-by-step explanation:

You start off with -3.9 x -5.7 which gives you a positive, and that positive is 22.23, then you multiply that number by 3, and you get 66.69.

7 0

3 years ago

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maks197457 [2]

What you must do in this case is to find the roots of the polynomial.
We have then:
x ^ 2 + 20x + 100 = 36
Rewriting:
x ^ 2 + 20x + 64 = 0
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x1 = -4
x2 = -16
Remember that the values of the roots are what make the polinome zero
Answer:
x1 = -4
x2 = -16

3 0

3 years ago

Evaluate the following double integral: xy dA D where the region D is the triangular region whose vertices are (0, 0), (0, 3), (

natulia [17]

Answer:

I= 84

Step-by-step explanation:

for

$I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{}_{} \int\limits^{}_D {x*y} \, dx*dy$

since D is the rectangle such that 0<x<3 , 0<y<3

$I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{3}_{0} \int\limits^{3}_{0} {x*y} \, dx*dy = \int\limits^{3}_{0} {x} \, dx\int\limits^{3}_{0} {y} \, dy = x^{2} /2*y^{2} /2 = (3^{2} /2 - 0^{2} /2)* (3^{2} /2 - 0^{2} /2) = 3^{4} /4 = 81/4$