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Ber [7]
3 years ago
5

Work our the size of angle x.

Mathematics
1 answer:
ANEK [815]3 years ago
8 0

Answer:

13°

Step-by-step explanation:

180-98=82

82+85=167

180-167= 13

x= 13°

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Point Q is the image of Q (-5,1) under a translation by 6 units to the right and 2 units down.
Alex787 [66]

Answer:

(1,-1)

Step-by-step explanation:

The graph is being shifted 6 units to the right which affects the x-value and it is also being affected 2 units down which affects the y-value.

6 0
3 years ago
Joe had 9 apples he now has 1 how many did he keep.
Firlakuza [10]
I hope this helps you



9-?=1


?=8
7 0
3 years ago
Read 2 more answers
I need to try and find the area.
Drupady [299]

Answer:

The answer is 32 square yards.

Step-by-step explanation: The formula for the area of a triangle is 1/2(b*h), which is one-half of the base multiplied by the height. 8 times 8 is 64. 64 divided by 2 is 32.

5 0
2 years ago
Let vector F = (6 x^2 y + 2 y^3 + 4 e^x) i + (7 e^{y^2} + 54 x) j . Consider the line integral of vector F around the circle of
balu736 [363]

Denote the circle of radius a by C. C is simple and closed, so by Green's theorem the line integral reduces to a double integral over the interior of C (call it D):

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_C(6x^2y+2y^3+4e^x)\,\mathrm dx+(7e^{y^2}+54x)\,\mathrm dy

=\displaystyle\iint_D\left(\frac{\partial(7e^{y^2}+54x)}{\partial x}-\frac{\partial(6x^2y+2y^3+4e^x)}{\partial y}\right)\,\mathrm dx\,\mathrm dy

=\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy

D is a circle of radius a, so we can write the double integral in polar coordinates as

\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^a(54-6r^2)r\,\mathrm dr\,\mathrm d\theta

a. For a=1, we have

\displaystyle\int_0^{2\pi}\int_0^1(54-6r^2)r\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^1(54r-6r^3)\,\mathrm dr=\boxed{51\pi}

b. Let I(a) denote the integral with unknown parameter a,

I(a)=12\pi\int_0^a(9r-r^3)\,\mathrm dr\,\mathrm d\theta

By the fundamental theorem of calculus,

I'(a)=12\pi(9a-a^3)

I(a) has critical points when

12\pi(9a-a^3)=12\pi a(9-a^2)=0\implies a=0,a=\pm3

If a=0, then line integral is 0, so we ignore that critical point. For the other two, we would find I(\pm3)=243\pi.

8 0
3 years ago
Thirty-five of the one hundred eighteen students who
Dmitriy789 [7]

Subtract the number that got an A from total students:

118 - 35 = 83

83 students did not get an A

7 0
3 years ago
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