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Luba_88 [7]
2 years ago
12

5/9+ (-4/5)

Mathematics
1 answer:
Virty [35]2 years ago
3 0
-11/45 is the correct choice
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Write the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 in this order and insert ‘+ ‘or ‘–' between them to get the result (a) 5 (b) –3
Liula [17]

Answer:

5

Step-by-step explanation:

we can write the digits in following form

0+1-2+3-4+5-6+7-8+9

= -1 - 1 -1 -1 +9

=-2-2+9

= -4+9

= 5

7 0
3 years ago
Answers? can someone help me please
vovangra [49]
I pretty sure the form is 2, and she bought 10 books, 5b-7=58
7 0
4 years ago
Read 2 more answers
Question Multiple Choice Worth 1 points)
just olya [345]

Answer:

Step-by-step explanation:

Givens

Area of base the right triangular prism = Area base of the cylinder

The heights of both are 6.

The general formula for a regular solid is V = B * height

The answer is that the volumes are the same

5 0
2 years ago
June has $1.95 in dimes and nickels. She has a total of 28 coins. How many of each type of coin does she have.
Verdich [7]

A dime = 0.10

A nickel = 0.05

N + D = 28 coins

Rewrite this as N = 28 -D


0.10D + 0.05N = 1.95

Replace N withe the rewritten equation above:

0.10D + 0.05(28-D) = 1.95

Use the Distributive Property:

0.10D + 1.4 - 0.05D = 1.94

Add the like terms:

0.05D + 1.4 = 1.95

Subtract 1.4 from both sides:

0.05D = 0.55

Divide both sides by 0.05

D = 0.55 / 0.05

D = 11

Nickels = 28 - 11 = 17


There are 11 dimes and 17 nickels.


3 0
4 years ago
Read 2 more answers
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
3 years ago
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