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3 years ago

What Value of m will make this equation true? 3 (m - 5) + 8 = 7 - 1/2 (4m - 11)

Mathematics

1 answer:

KatRina [158]3 years ago

6 0

Answer:

m = 7.1

Step-by-step explanation:

3 (m -5) + 8 = 7 -1/2 (4m - 11)

So, what I did first was distribute the values in the parenthesis to get,

3m - 15 + 8 = 7 -2m -5.5

Now that we have the parenthesis taken care of we can do the simpler math,

3m - 23 = 7 -2m - 5.5

I just added the 15 and 8, so now I move the -5.5 to the opposite side by adding.

3m - 23 = 7 -2m -5.5

+5.5 +5.5

3m - 28.5 = 7 - 2m

Here we can do the same with the -2m.

3m - 28.5 = 7 - 2m

+2m +2m

5m - 28.5 = 7

To get rid of the -28.5 I added it to the 7 getting an answer of,

5m - 28.5 = 7

+ 28.5 +28.5

5m = 35.5

Finally, divide 5m and 35.5 both by 5.

5m/5 = m

35.5/5 = 7.1

Answer: m=7.1

Sorry it's really long, but I hope this helps! Have a great day!

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Can somebody teach me how to do this?

galina1969 [7]

Ratios and fractions are interchangeable (for the most part). A fraction is a type of ratio. And we've all worked with fractions before, so that's what we'll do here.

I can say that the ratio 3:1 is the same as the fraction 3 / 1. Using that fraction, it is fairly easy to find equivalent fractions.

3 / 1 = 6 / 2 = 9 / 3 = 30 / 10

...And so on

Another example, 5:6 = 5 / 6.

5 / 6 = 10 / 12 = 15 / 18 = 30 / 36

To find equivalent fractions, all you need to do is multiply both the numerator and denominator by the same number. As long as that number is the same (and your multiplication is correct), then the new fraction will be equivalent to the original.

Hope this helps!

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2 years ago

Consider the differential equation x^2 y''-xy'-3y=0. If y1=x3 is one solution use redution of order formula to find a second lin

Anastasy [175]

Suppose $y_2(x)=y_1(x)v(x)$ is another solution. Then

$\begin{cases}y_2=vx^3\\{y_2}'=v'x^3+3vx^2//{y_2}''=v''x^3+6v'x^2+6vx\end{cases}$

Substituting these derivatives into the ODE gives

$x^2(v''x^3+6v'x^2+6vx)-x(v'x^3+3vx^2)-3vx^3=0$

$x^5v''+5x^4v'=0$

Let $u(x)=v'(x)$ , so that

$\begin{cases}u=v'\\u'=v''\end{cases}$

Then the ODE becomes

$x^5u'+5x^4u=0$

and we can condense the left hand side as a derivative of a product,

$\dfrac{\mathrm d}{\mathrm dx}[x^5u]=0$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}[x^5u]\,\mathrm dx=C$

$x^5u=C\implies u=Cx^{-5}$

Solve for $v$ :

$v'=Cx^{-5}\implies v=-\dfrac{C_1}4x^{-4}+C_2$

Solve for $y_2$ :

$\dfrac{y_2}{x^3}=-\dfrac{C_1}4x^{-4}+C_2\implies y_2=C_2x^3-\dfrac{C_1}{4x}$

So another linearly independent solution is $y_2=\dfrac1x$ .

3 0

3 years ago

Can somebody please help me

seraphim [82]

Answer:

3) 36

2) 55

Step-by-step explanation:

We have to place the numbers in order from least to greatest.

3) 23, 26, 26, 32, 32, 36, 50, 52, 52, 52, 59,

4) 40, 42, 49, 54, 62, 67, 67, 70, 95

The more is the the difference between the lowest and highest values.

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6 0

3 years ago

Find the length to the nearest centimeter of the diagonal of a square 30 cm. on a side

tiny-mole [99]

Answer:

42 cm.

Step-by-step explanation:

Please find the attachment.

Let x be the length of diagonal of the square.

We have been given that length of each side of a square is 30 cm. We are asked to find the length of the diagonal of square to the nearest centimeter.

We can see from our diagram that triangle AC is the diagonal of our square.

Since all the interior angles of a square are right angles or equal to 90 degrees, so we will use Pythagoras theorem to find the length of diagonal.

$AC^2=AD^2+DC^2$