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zhenek [66]
3 years ago
5

Can anyone please help me on this?

Mathematics
1 answer:
Elena L [17]3 years ago
3 0

Answer:

b

Step-by-step explanation:

ive done this beffore im a pro

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In triangle ABC, (angle A) is a right angle and (m angle B) = 45deg.
Ratling [72]
We can use the ratios for special triangles (see the attachment below). We'll be using the 45-45-90 triangle. We notice that side BC is equal to x \sqrt{2}, and side AC (x using the special right triangle) is equal to 16. We can therefore say that side BC is equal to 16 \sqrt{2} ft

:)

4 0
3 years ago
Read 2 more answers
Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ
atroni [7]

Answer: y=Ce^(^3^t^{^9}^)

Step-by-step explanation:

Beginning with the first differential equation:

\frac{dy}{dt} =27t^8y

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

\frac{1}{y} \frac{dy}{dt} =27t^8

Multiply both sides by 'dt' to get:

\frac{1}{y}dy =27t^8dt

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt

ln(y)=27(\frac{1}{9} t^9)+C

ln(y)=3t^9+C

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)

y=e^(^3^t^{^9} ^+^C^)

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

y=e^(^3^t^{^9}^)e^C

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

y=Ce^(^3^t^{^9}^)

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))

\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)

\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8

Now check if the derivative equals the right side of the original differential equation:

(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)

Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

7 0
2 years ago
Who is the partial product of 28 and 42
Lana71 [14]

The partial products are: 42 x 28 = (40 + 2) x (20 + 8) = 40 x 20 + 40 x 8 + 2 x 20 + 2 x 8
= 800 + 320 + 40 + 16 Hope this helped.
6 0
3 years ago
I NEED HELP ASAP!!! I'LL GIVE BRAINLIEST TO THE CORRECT ANSWER!!!
icang [17]

Answer:

v=18

u=4

Step-by-step explanation:

v/3=6

v=18

2u+2=5u-10

3u=12

u=4

8 0
3 years ago
Read 2 more answers
a portion of 7200 is invested at 5 1/2 percent interest, and the rest is invested at 5 percent interest. if the yearly income fr
lana [24]

Answer:

A 5.5% = 5400 B 5% = 1800

Step-by-step explanation:

A + B = 7200

A * 0.055 + B * 0.05 = $387

A = 7200 - B

(7200 - B)*0.055 + B*0.05 = 387

396 - 0.055B + 0.05B = 387

396 - 0.005B = 387

0.005B = 9

B = 1800

A = 7200 - B = 7200 - 1800

A = 5400

5 0
3 years ago
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