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slava [35]
3 years ago
8

Mariah went to the grocery store and purchased m cartons of milk and s bags of sugar. The milk was $2.39 per carton and the suga

r was $1.99 per bag. She also purchased c candy bars for $2.35 each. On the way out of the store she donated $4 to a local charity.
Evaluate the following expression to determine how much Mariah spent at the grocery store:

2m + s + 3c + 4

Explain how to evaluate the expression and determine how much money Mariah spen
Mathematics
1 answer:
Solnce55 [7]3 years ago
8 0
Your answer would be 10.73 you would evaluate by going 2.39m + 1.99pb + 2.35c + 4lc = 10.73
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BRAINLIEST ANSWER PLUS EXTRA POINTS! Two planes leaves an airport at the same time, one flying east one flying west. The eastbou
nirvana33 [79]

Equation:

distance + distance = 2250 miles

3x + 3(x-110) = 2250

6x - 330 = 2250

6x = 2580

x = 430 mph (rate of the westbound plane)

x-110 = 320 mph (rate of the eastbound plane)


PLS MARK BRAINLIEST!!!

5 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
3 years ago
Plz help me on this question
zhuklara [117]
I cant see it please send another pic of it. closer to the screen.
5 0
4 years ago
Write the numbers 23,643 and 23,987 so they line up by place value. Explain how to line them up
weqwewe [10]

Consider the numbers 23,643 and 23,987.

Since both the given numbers have same digits at ten thousands and thousands place.

So, we can'not compare the numbers on the basis of ten thousands and thousands digit.

So, let us consider the number 23,643

Consider the place value of digit at the hundreds place. Since, the digit at hundreds place = 6.

The place value of 6 = 600

Now, let us consider the number 23,987

Consider the place value of digit at the hundreds place. Since, the digit at hundreds place = 9.

Te place value of 9 = 900

Since, 900 > 600

So, 23,643 > 23,987

Therefore, the number 23,987 is greater than the number 23,643.

4 0
3 years ago
i a man even number beetween 50 and 99 i am a multiple of 9 my tens digit is 5 more than my units digit
vodomira [7]
Hello,
multiple of 9 betheen 50 and 99 are :54,63,72,81,99
and 54,72  are even.
5-4=1 and not 5
7-2=5 ok

72 is the number searched.
6 0
4 years ago
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