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Sati [7]
3 years ago
11

At least one of the numbers must be 0. 1. none 2. Hypothesis 3. Conclusion

Mathematics
1 answer:
iragen [17]3 years ago
5 0

Answer:

The answer is 1.) None

Step-by-step explanation:

Hope this helps you :)

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Factor. 2v^2-11v-63
Tatiana [17]

Answer:

  (2v +7)(v -9)

Step-by-step explanation:

Factors of 2·(-63) = -126 that have a sum of -11 are 7 and -18, so we can rewrite the middle term to -18v+7v. Then we can factor by grouping.

  2v^2 -18v +7v -63 = 2v(v -9) +7(v -9)

  = (2v +7)(v -9)

5 0
3 years ago
What is tan c as a simplified fraction and what is sin b as a simplified fraction?
aniked [119]

Answer: SOH CAH TOA Tan C= 11/22= 1/2 So Tan C= 1/2

Sin B= x/17


Step-by-step explanation:

SOH CAH TOA = S=o/h C=a/h T=o/a.

I love trig ratios!

Hope this helped.

3 0
3 years ago
The speed limit for Stadium Drive is 45 miles per hour. Generate at least 3 different interpretations of the rate 45 miles per h
bija089 [108]

Step-by-step explanation:

The solution to this problem requires that we find other units besides the given one

1. ft/sec

2. km/h

3.m/s

1. converting to ft/sec

1 mile is 5280 ft

=\frac{45mile}{hour}* \frac{5280feet}{1 mile} *\frac{1 hours}{3600 secs} \\\\=66 ft/sec

2. converting to km/h

1 mile is 1.60934

=\frac{45mile}{hour}* \frac{1.60934km}{1 mile} *\frac{1 hours}{1hour} \\\\=72.42km/h

3. converting to m/s

1 mile is 1609.34

=\frac{45mile}{hour}* \frac{1609.34km}{1 mile} *\frac{1 hours}{3600 sec} \\\\=20.11m/s

6 0
3 years ago
HELLPPP PLZZZ ASAP
Alla [95]

Answer: idk how to explain but I learned about tectonic plates and from this picture the boundaries are sliding past each other

Step-by-step explan

6 0
3 years ago
) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra
artcher [175]

Answer:

y(t)=2e^{3t}(2-5t)

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2  

Using the theorem of the Laplace transform for derivatives, we know that:

\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)

Replacing the initial values y(0)=4, y′(0)=2 we obtain

\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4

and our differential equation (*) gets transformed in the algebraic equation

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0

Solving for Y(s) we get

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}

Now, we brake down the rational expression of Y(s) into partial fractions

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here  

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}

and

\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}

we know that

\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}

and for the first translation property of the inverse Laplace transform

\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}

and the solution of our differential equation is

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}

5 0
3 years ago
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