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sergiy2304 [10]
3 years ago
8

Leila works mowing lawns and babysitting. She earns $8.30 an hour for mowing and $7.30 an hour for babysitting. How much will sh

e earn for 6 hours of
mowing and 4 hours of babysitting?
Mathematics
1 answer:
Mamont248 [21]3 years ago
5 0

Answer:

$79 total earnings

Step-by-step explanation:

Here, earnings = (hourly rate)(number of hours):

                          = ($8.30/hr)(6 hrs) + ($7.30/hr)(4 hrs), or

Total earnings =  $49.80               +  $29.20

    "          "         = $79.00

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Ten times the square of a non zero number is equal to ninety times the number. what is the number?
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I think the answer is 9! hope this helps!!
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3 years ago
I only need the Evens ;-;
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Supplementary is when two angles add up to 180
Complementary is when two angles add up to 90
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21. Supplementary

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23. Angle 2 is 141

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7 0
3 years ago
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Marina86 [1]

Answer:

B.

Step-by-step explanation:

b. because its the L in the cearl bowl

7 0
3 years ago
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Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m &gt; 1 so that th
jonny [76]

Answer:

The answer is \sqrt{\frac{6}{5}}

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy

Solving:

\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy

[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.

Replacing the limits:

6*\frac{1}{3} =2

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ \frac{1}{m}

Solving the double integral with these new limits we have:

\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy

This part is a little bit tricky so let's solve the integral first for dy:

\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx

Replacing the limits:

\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx

Solving now for dx:

[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.

Replacing the limits:

\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}

As I mentioned before, this volume is equal to 1, hence:

\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }

3 0
3 years ago
Find the GCF (greatest common factor) of the following terms.<br> {5x2y2,25x3y2,x2y2}
HACTEHA [7]

Answer:

5x2y2 is the GCF common factor of the following terms.

7 0
2 years ago
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