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kirza4 [7]
3 years ago
13

The lengths of the legs of a right triangle are 8 inches and 9 inches. What is the length of the hypotenuse? Round to the neares

t inch. Enter your answer in the box.
Mathematics
1 answer:
patriot [66]3 years ago
6 0
Hyp^2 = 8^2 + 9^2
hyp^2 = 64 + 81
hyp^2 = 145
hot = 12 inch
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Solve 2 log(x) + log(5) = log (80) using the one-to-one property.
vodomira [7]

x=4

Step-by-step explanation:

2log(x)+log(5)=log(80)

log(x^2)+log(5)=log(80)

log(5x^2)=log(80)

5x^2 =80

x^2 =80/5

x^2=16 , taking square root both sides

x=√16

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5 0
3 years ago
9(8s - 5f) use f = 3 and s = 4
Savatey [412]

Answer:

153

Step-by-step explanation:

8x4=32

5x3=15

32-15=17

9x17=153

5 0
3 years ago
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How many x-intercepts does the graph of each quadratic equation have? Show your work.
Alex73 [517]

Answer:

Step-by-step explanation:

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If the discriminant is positive, we  have two unequal real roots.

#51:  8v^2 - 12v + 9:  the discriminant is (-12)^2 - 4(8)(9) = -144.  we have two unequal, complex roots

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4 0
3 years ago
Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.
fredd [130]

\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}

\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}

so the ODE is indeed exact and there is a solution of the form F(x,y)=C. We have

\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)

\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)

f'(y)=0\implies f(y)=C

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With y(1)=2, we have

8+\ln9=C

so

\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}

8 0
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PLEASE ITS DO TOMORROW
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Here you go :) !!!!!!!!

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