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4 years ago

20. what us the area of the figure:

Mathematics

1 answer:

Varvara68 [4.7K]4 years ago

8 0

If the angle in B is 45º, that means the angle in A is also going to be 45º, considering it's a right triangle, so AC and BC are they have the same length.
Then using Pythagoras, you'll get that $AC^{2} + BC^{2}$ equals $AB^{2}$ .
Now, you know that AC=BC and that AB=24.
So you'll get $2AC^{2} =24^2$ . You do the square root in both sides and you get that 2AC=24 and AC=12.

Now that you know that both AC and BC equal 12, you can find the area by just multiplying them and then diving them by 2. (The formula for the area of a triangle is half base multiplied by height, and in a right triangle, if a cathetus is the base, the other cathetus<span> will be the height)

And so the area is equal to 72.</span>

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The graph shows the function y equal 3x-1 what are the coordinates of the intercepts

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Answer: (0,-1) and (1/3,0)

Step-by-step explanation:

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3 years ago

3. Use the following table to answer questions (a – c).

sineoko [7]

Answer:

See below

Step-by-step explanation:

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3 years ago

Please help ASAP work and answer please please

tiny-mole [99]

Answers:

8. 12x+3= 27

move 3 to the other side , sign changes from +3 to -3

12x+3-3=27-3

12x= 24

divide by 12 for both sides

12x/12=24/12

x= 2

9. 6(y-10)=42

mutiply the bracket by 6

(6)(y)(6)(-10)= 6y-60

6y-60= 42

move -60 to the other side, sign changes from -60 to +60

6y-60+60=42+60

6y=102

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move +4x to the other side

9x-4x-2= 4x-4x+13

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11. 4/3y+5/2y= 1

find the common denominator for both of the fractions which is 6.

Mutiply by 2 for 4/3y .

4(2)/3(2)y=8/6y

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23/6y= 1

Mutiply both sides by 6/23

23/6y(6/23)= 1 (6/23)

Cross out 6 and 6 , divide by 6 and then becomes 1.

Cross out 23 and 23, divide by 23 and then becomes 1.

1*1*y=y

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4 years ago

) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle

Fynjy0 [20]

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Parameterize $S$ by

$\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)$

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$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv$

$=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}$

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4 years ago

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Step-by-step explanation:

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3 years ago