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kobusy [5.1K]
3 years ago
5

Help me pleaseeeeeeeee

Mathematics
2 answers:
kirill [66]3 years ago
8 0

Answer:

B. 3.45

Step-by-step explanation:

The radius is half the diameter

Scrat [10]3 years ago
4 0

Answer:

B 3.45

Step-by-step explanation:

the radius is always diameter/2

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Solve the system of equations. y = -5x + 24 y = 4x - 21 a. ( -5, -1) c. ( -1, 5) b. ( 5, -1) d. No solution
Svetlanka [38]

y = -5x + 24

y = 4x - 21

Since both of these equations are equal to Y, theyre equal to each other.

So we can make an equation with y = -5x + 24 in one side and y = 4x - 21 on the other.

-5x + 24 = 4x - 21

Now in order to get the value of x we need to isolate it in one side of the equation. We can do this by subtracting 24 from both sides of the equation:

-5x + 24 - 24 = 4x - 21 - 24

-5x = 4x - 45

Now we subtract 4x from both sides so the 4x shift to the other side

-5x - 4x = 4x - 4x - 45

-9x = -45

Finally divide both sides by -9 so x is by itself

(-9)÷(-9x) = -(45)÷(-9)

x = 5

Since we did all of this to BOTH sides of the equation, both sides are still equal to each other and the equation still is true.

Now apply x = 5 to either of the initial equations to find the value of Y

y = -5x + 24 or y = 4x - 21

(I'll do both but u only need one)

y = -5(5) + 24

y = -25 + 24

y = -1

y = 4(5) - 21

y = 20 - 21

y = -1

Either way, X is 5 and Y is -1

Answer (5, -1)

5 0
3 years ago
Please help me with the below question.
Snezhnost [94]

6a. By the convolution theorem,

L\{t^3\star e^{5t}\} = L\{t^3\} \times L\{e^{5t}\} = \dfrac6{s^4} \times \dfrac1{s-5} = \boxed{\dfrac5{s^4(s-5)}}

6b. Similarly,

L\{e^{3t}\star \cos(t)\} = L\{e^{3t}\} \times L\{\cos(t)\} = \dfrac1{s-3} \times \dfrac s{1+s^2} = \boxed{\dfrac s{(s-3)(s^2+1)}}

7. Take the Laplace transform of both sides, noting that the integral is the convolution of e^t and f(t).

\displaystyle f(t) = 3 - 4 \int_0^t e^\tau f(t - \tau) \, d\tau

\implies \displaystyle F(s) = \dfrac3s - 4 F(s) G(s)

where g(t) = e^t. Then G(s) = \frac1{s-1}, and

F(s) = \dfrac3s - \dfrac4{s-1} F(s) \implies F(s) = \dfrac{\frac3s}{\frac{s+3}{s-1}} = 3\dfrac{s-1}{s(s+3)}

We have the partial fraction decomposition,

\dfrac{s-1}{s(s+3)} = \dfrac13 \left(-\dfrac1s + \dfrac4{s+3}\right)

Then we can easily compute the inverse transform to solve for f(t) :

F(s) = -\dfrac1s + \dfrac4{s+3}

\implies \boxed{f(t) = -1 + 4e^{-3t}}

6 0
2 years ago
Use the distributive property to remove the parentheses.<br> -3(-6y +3w-2)
LekaFEV [45]

Answer:

18y-9w+6

Step-by-step explanation:

-3*-6y=18y

-3*3w=-9w

-3-2=6

so, 18y-9w+6

3 0
2 years ago
Read 2 more answers
Need help ASAP! Tysm
UNO [17]

Answer:

The answer is B) -29.

Step-by-step explanation:

3 0
4 years ago
Help me solve this please. I don't know the formula for this.​
den301095 [7]

Answer

48,000

Step-by-step explanation:

base(x)

1,200(40)

8 0
4 years ago
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