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3 years ago

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The set of numbers below were listed on me.Carter's homework page. Order the numbers from less to greatest (10, -2, |-16 |,-7, |

3 | .14 , |-9 | . 15, -1, 0

1 answer:

ANTONII [103]3 years ago

6 0

Answer:-2,-1,0,|.15|,.15, |3|, |-7|, |-9|, 10, | -16 |

Step-by-step explanation:

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The sum of three consecutive multiples of 4 is 444. Find the multiples.<br> PLEASE ANS FAST PLS

SOVA2 [1]

Answer:

144,148,152

Step-by-step explanation:

Here we will have, 4x + 4(x+1) + 4(x+2) = 444

=> x + (x+1) + (x+2) = 111 (dividing both sides by 4)

=>x + x+1 + x+2 =111

=> 3x + 3 = 111

=> 3x = 108

=> x = 36.

So, the three desired numbers 4x, 4(x+1) and 4(x+2) are,

4*36, 4*(36+1) and 4*(36+2)

That means the numbers are, 144, 148 and 152.

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3 years ago

Can someone explain to me what it means to be marked as branliest? & what it does ?

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3 years ago

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Gala2k [10]

<h3>Total snow fall for three days is 8.75 inches</h3>

<em><u>Solution:</u></em>

Given that,

Snow fell over the past three days

$First\ day = 3\frac{2}{4}\ inches = \frac{14}{4}\ inches\\\\Second\ day = 3\frac{2}{4}\ inches = \frac{14}{4}\ inches\\\\Third\ day = 1\frac{3}{4}\ inches = \frac{7}{4}\ inches$

<em><u>What was the total snowfall for the three days?</u></em>

Total snowfall = first day + second day + third day

$Total\ snowfall = \frac{14}{4} + \frac{14}{4} + \frac{7}{4}\\\\Total\ snowfall = \frac{14 + 14 + 7}{4}\\\\Total\ snowfall = \frac{35}{4} \text{ or } 8.75\ inches$

Thus total snow fall for three days is 8.75 inches

5 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

4 years ago

Right triangle ABC has three sides with lengths AB= 119, BC = 169, CA= 120. Find the value of cos C. hint: draw and label the tr

GenaCL600 [577]

Answer:

$\cos(C)=\frac{120}{169}$

Step-by-step explanation:

The triangle is shown in the attachment.

Recall that;

$\cos(C)=\frac{Adjacent}{Hypotenuse}$

This implies that;

$\cos(C)=\frac{|AC|}{|BC|}$

We substitute the values into the ratio to obtain

$\cos(C)=\frac{120}{169}$

8 0

3 years ago