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9966 [12]
2 years ago
6

What is the value of x in the equation 8x – 2y = 48, when y = 4? 6 7 14 48.

Mathematics
1 answer:
olga nikolaevna [1]2 years ago
7 0

Answer: 7

Step-by-step explanation:

8x-2y=48

If y=4 then put the value of y in this equation

8x-2(4)=48

8x-8=48

We take common 8 both the side

8[x-1]=8[6]

Now cancel it then

X-1=6

X=6+1

X=7

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9. Aaron has a lemonade business. If he sells each glass of lemonade for $2.75, and he sold 11
marin [14]

Answer:

$27.75

Step-by-step explanation:

Multiply $2.25 by 11 to get the answer

4 0
3 years ago
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What can you do when the unit prices for two comparable items are listed using different units of measure?
Over [174]
When you have two unit prices for two comparable items, you have to convert one of the prices to the other one.
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3 years ago
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A random representative survey of 100 seventh-graders in Washington Middle School shows that 20 students read 5 or more books ov
DaniilM [7]

20 out of 100 read 5 or more.

20/100 = 0.2

Now multiply the total students by 0..2:

275 x 0.2 = 55

55 people read 5 or more books.

8 0
2 years ago
Set up the integral that represents the arc length of the curve f(x) = ln(x) + 5 on [1, 3], and then use Simpson's Rule with n =
marta [7]

Answer:

The integral for the arc of length is:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx

By using Simpon’s rule we get: 1.5355453

And using technology we get:  2.3020

The approximation is about 33% smaller than the exact result.

Explanation:

The formula for the length of arc of the function f(x) in the interval [a,b] is:

\displaystyle\int_a^b \sqrt{1+[f'(x)]^2}dx

We need the derivative of the function:

f'(x)=\frac{1}{x}

And we need it squared:

[f'(x)]^2=\frac{1}{x^2}

Then the integral is:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx

Now, the Simposn’s rule with n=4 is:

\displaystyle\int_a^b g(x)}dx\approx\frac{\Delta x}{3}\left( g(a)+4g(a+\Delta x)+2g(a+2\Delta x) +4g(a+3\Delta x)+g(b) \right)

In this problem:

a=1,b=3,n=4, \displaystyle\Delta x=\frac{b-a}{n}=\frac{2}{4}=\frac{1}{2},g(x)= \sqrt{1+\frac{1}{x^2}}

So, the Simposn’s rule formula becomes:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\\\approx \frac{\frac{1}{3}}{3}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{1}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(1+\frac{2}{2}\right)^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{3}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)

Then simplifying a bit:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx \approx \frac{1}{9}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(\frac{3}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(2\right)^2}} +4\sqrt{1+\frac{1}{\left(\frac{5}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)

Then we just do those computations and we finally get the approximation via Simposn's rule:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\approx 1.5355453

While when we do the integral by using technology we get: 2.3020.

The approximation with Simpon’s rule is close but about 33% smaller:

\displaystyle\frac{2.3020-1.5355453}{2.3020}\cdot100\%\approx 33\%

8 0
3 years ago
2(4-6+1)=-5+ 8:4 =<br> What’s the answer
allochka39001 [22]

Step-by-step explanation:

First you do the one in the parentheses: 4 - 6 = -2

Then you got the answer: -2

Then you add -2 + 1 = -1

You got the answer: -1

8 0
3 years ago
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