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damaskus [11]
3 years ago
15

Find the length of the hypotenuse.​

Mathematics
1 answer:
IRISSAK [1]3 years ago
8 0

Answer:

6

Step-by-step explanation:

Using the sine ratio in the right triangle

let x = hypotenuse and sin45° = \frac{\sqrt{2} }{2}, then

sin45° = \frac{opposite}{hypotenuse} = \frac{3\sqrt{2} }{x}

and

\frac{3\sqrt{2} }{x} = \frac{\sqrt{2} }{2} ( cross- multiply )

\sqrt{2} × x = 6\sqrt{2}

Divide both sides by \sqrt{2}

x = 6

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Each of the six letters stands for a different number, 1 through 9. Using the three clues, can you deduce the number represented
-BARSIC- [3]

Answer:

A = 9

B = 2

C = 5

D = 1

E = 8

F = 7

Step-by-step explanation:

A 9 B 2 C 5 = 16

D 1  E 8 F 7 = 16

10   10    12

BxC (2x5) = 10

A+D (9+1) = 10

4 0
4 years ago
The area of a square with side a is S. What is the formula for dependence of S on a?
lawyer [7]

Answer:

S=a^2

Step-by-step explanation:

Note all sides of a square are equal, and the formula for finding the area is length x width of which the length=a and the with =a. So S=a^2.

4 0
3 years ago
Need Help on math question <br><br>​
Nikitich [7]

Answer:

36 yards

Step-by-step explanation:

The triangle is a 3-4-5 right triangle with a scaling factor of 12. No complicated Pythagorean theorem has to be used in this case.

60/12 = 5

48/12 = 4

3 is the only base number remaining.

3(12) =36

4 0
3 years ago
PLEASE HELP ITS DUE TODAY PLEASE HELP ASAP 15 POINTS PLEASE HELP​
ludmilkaskok [199]

Answer:

210 cm³

Step-by-step explanation:

volume = W x L x H

(where W = width, L = length and H = height)

By inspection of the diagram,

W = 6 cm

L = 7 cm

H = 5 cm

Therefore, volume = 6 x 7 x 5 = 210 cm³

6 0
3 years ago
Read 2 more answers
A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min
Dmitry [639]

At any time t (min), the volume of solution in the tank is

500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}

If A(t) is the amount of salt in the tank at any time t, then the solution has a concentration of \dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}.

The net rate of change of the amount of salt in the solution, A'(t), is the difference between the amount flowing in and the amount getting pumped out:

A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)

Dropping the units and simplifying, we get the linear ODE

A'=10+5\sin\dfrac t4-\dfrac A{10}

10A'+A=100+50\sin\dfrac t4

Multiplying both sides by e^{10t} allows us to identify the left side as a derivative of a product:

10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}

\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}

e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt

Integrate and divide both sides by e^{10t} to get

A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}

The tanks starts off with 30 lb of salt, so A(0)=30 and we can solve for C to get a particular solution of

A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}

6 0
4 years ago
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