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bearhunter [10]
2 years ago
15

5. The area of the trapezoid is 75 square inches. Find the height of the trapezoid. And it’s two bases are 10 and 15.

Mathematics
1 answer:
GaryK [48]2 years ago
7 0
6

6
h/2(sum of bases) =75=area
h=3*2


Mark brainliest please
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Zoey inflated 24 balloons for the high school dance. If Zoey inflated 15% of the total number of balloons at the dance, how many
Eva8 [605]

Answer:

Total number of balloons= 160

Step-by-step explanation:

Giving the following information:

Number of balloons inflated by Zoey= 24

Proportion of balloons inflated= 15%

<u>To calculate the total number of balloons, we need to use the following formula:</u>

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What is the greatest common factor (GCF) of the numerator and denominator in the rational expression below? 6x-18/x^2-5x+6
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Solve the following differential equations or initial value problems. In part (a), leave your answer in implicit form. For parts
shepuryov [24]

Answer:

(a) (y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) y = arctan(t(lnt - 1) + C)

(c) y = -1/ln|0.09(t + 1)²/t|

Step-by-step explanation:

(a) dy/dt = (t^2 + 7)/(y^4 - 4y^3)

Separate the variables

(y^4 - 4y^3)dy = (t^2 + 7)dt

Integrate both sides

(y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) dy/dt = (cos²y)lnt

Separate the variables

dy/cos²y = lnt dt

Integrate both sides

tany = t(lnt - 1) + C

y = arctan(t(lnt - 1) + C)

(c) (t² + t) dy/dt + y² = ty², y(1) = -1

(t² + t) dy/dt = ty² - y²

(t² + t) dy/dt = y²(t - 1)

(t² + t)/(t - 1)dy/dt = y²

Separating the variables

(t - 1)dt/(t² + t) = dy/y²

tdt/(t² + t) - dt/(t² + t) = dy/y²

dt/(t + 1) - dt/(t(t + 1)) = dy/y²

dt/(t + 1) - dt/t + dt/(t + 1) = dy/y²

Integrate both sides

ln(t + 1) - lnt + ln(t + 1) + lnC = -1/y

2ln(t + 1) - lnt + lnC = -1/y

ln|C(t + 1)²/t| = -1/y

y = -1/ln|C(t + 1)²/t|

Apply y(1) = -1

-1 = ln|C(1 + 1)²/1|

-1 = ln(4C)

4C = e^(-1)

C = (1/4)e^(-1) ≈ 0.09

y = -1/ln|0.09(t + 1)²/t|

8 0
3 years ago
Please answer right ill give brainliest to the right answerrrrrrr
morpeh [17]
It’s would be 24 photos per day
3 0
2 years ago
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