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2 years ago

7 (m+5) = 21 as a step by step problem?

Mathematics

2 answers:

Sever21 [200]2 years ago

8 0

Answer:

m=-2

Step-by-step explanation:

You want to isolate the variable on one side (m).

Divide each side by 7.

m+5=3

Subtract 5 from both sides

m=-2

lina2011 [118]2 years ago

6 0

Answer:

m = -2.

Step-by-step explanation:

First divide both sides by 7; m + 5 = 3.

Then subtract 5 from both sides; m = -2.

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X + y = 152,<br><br> 8.5x + 12y = 1,656<br><br> How many hats were sold?

Elodia [21]

Answer:

x = 48 and y = 104

Step-by-step explanation:

Given equations are:

$x+y = 152\\8.5x+12y=1656$

From equation 1:

$x = 152-y$

Putting the value of y in equation 2

$8.5(152-y)+12y = 1656\\1292-8.5y+12y = 1656\\3.5y+1292 = 1656\\3.5y = 1656-1292\\3.5y = 364\\\frac{3.5y}{3.5} = \frac{364}{3.5}\\y = 104$

Now we have to put the value of y in one of the equation to find the value of x

Putting y = 104 in the first equation

$x+y = 152\\x+ 104 = 152\\x = 152-104\\x = 48$

Hence,

The solution of the system of equations is x = 48 and y = 104

The value of variable which was assumed for number of hats, is the total number of hats.

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

Help me plz it’s ASAP I need it

Lady_Fox [76]

Answer:

Step-by-step explanation:

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5 0

3 years ago

What happens when an inequality expression is divided by the same number. Does the sign stay the same each time or change direct

irga5000 [103]

The sign only changes when divided or multiplied by a negative number.

7 0

3 years ago

Read 2 more answers

Solve 25n=2×25 for n. You can use a related equation. n = <img src="https://tex.z-dn.net/?f=%5Cfrac%7B2%2A25%7D%7B25%7D" id="Tex

Archy [21]

Answer:

n=2

Step-by-step explanation:

25n=2×25

Divide each side by 25

25n/25=2×25/25

n = 2 * 25/25

n =2

3 0

3 years ago