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Ilya [14]
2 years ago
12

How old am I if 2 times my age increased by 244 is 400?

Mathematics
2 answers:
ikadub [295]2 years ago
8 0

Answer:

78

Step-by-step explanation:

400-244=156

156/2=78

MrRissso [65]2 years ago
7 0

Answer:

78 years old

Step-by-step explanation:

400 − 2 x = 244

400 − 244 = 2 x

156 = 2 x

so the answer would be 78 years old...

Hope that was helpful.Thank you!!!

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One number is seven more than a second number. Twice the first is 2 less than 3 times the second. Find the numbers.
Keith_Richards [23]

Step-by-step explanation:

Linear equations. Make 2 new equations from the question. Name the first number as x and the second as y...

4 0
3 years ago
Evaluate the double integral.
Fynjy0 [20]

Answer:

\iint_D 8y^2 \ dA = \dfrac{88}{3}

Step-by-step explanation:

The equation of the line through the point (x_o,y_o) & (x_1,y_1) can be represented by:

y-y_o = m(x - x_o)

Making m the subject;

m = \dfrac{y_1 - y_0}{x_1-x_0}

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

m= \dfrac{2-1}{1-0}

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

m = \dfrac{1-2}{4-1}

m = \dfrac{-1}{3}

∴

y-2 = -\dfrac{1}{3}(x-1)

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg)   dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ([y^2(-4y+8)] \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( -4y^3+8y^2 \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1

\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1

\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{-45+56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{11}{3}\bigg]

\iint_D 8y^2 \ dA = \dfrac{88}{3}

4 0
2 years ago
Find the area of the parallelogram. Please this is due today. Thank you.
scZoUnD [109]
894 mi^2

as 24 x 37.25 = 894
4 0
3 years ago
A triangle has 2 sides that are 9 inches and the bottom is 6 inches what is the height?
Elanso [62]

The height of the isosceles triangle is 8.49 inches.

<h3>How to find the height of the triangle?</h3>

Here we have a triangle such that two of the sides measure 9 inches, and the base measures 6 inches.

So this is an isosceles triangle.

We can divide the isosceles triangle into two smaller right triangles, such that the side that measures 9 inches is the hypotenuse, the base is 3 inches, and the height of the isosceles triangle is the other cathetus.

By Pythagorean's theorem, we can write:

(9in)^2 = (3 in)^2 + h^2

Where h is the height that we are trying to find.

Solving that for h we get:

h = √( (9 in)^2 - (3in)^2) = 8.49 inches.

We conclude that the height of the isosceles triangle is 8.49 inches.

If you want to learn more about triangles:

brainly.com/question/2217700

#SPJ1

8 0
2 years ago
A school bought 32 new desks. Each desk cost $24. Estimate how much the school spent on the new desks.
Pavlova-9 [17]

Estimate:

32 x 24

30 x 25

$750



8 0
3 years ago
Read 2 more answers
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