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Mademuasel [1]
2 years ago
6

The sum of 237 and six times the opposite of a number is 405. What is the number?

Mathematics
2 answers:
AleksAgata [21]2 years ago
5 0
Look in the file below for the answer

Alona [7]2 years ago
4 0
237+6z = 405

405-237=168

6z = 168
168/6 = 28
z=28

good days.
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Solve the inequality. 4 + 9x > 8 − 3x
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Answer:

x > 1/3

Step-by-step explanation:

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The diameter of a circle if five times the length of a recatngle. The length of a rectangle is three times the width of the rect
Leno4ka [110]

Answer:

706.5 ft²

Step-by-step explanation:

A. Width of the rectangle

The formula for the area of a rectangle is

A = lw

But, l = 3w, so,

A = (3w)w

= 3w² = 12

      w² = 4         Divided each side by 3

      w   = 2 ft     Took the square root of each side

B. Length of rectangle

l = 3w = 6 ft

C. Diameter of circle

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D. Radius of circle

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8 0
3 years ago
Find lim x→3 sqrt 2x+3-sqrt 3x/ x^2-3x. you must show your work or explain your work in words plsss I need help
sineoko [7]

I'm assuming the limit is supposed to be

\displaystyle\lim_{x\to3}\frac{\sqrt{2x+3}-\sqrt{3x}}{x^2-3x}

Multiply the numerator by its conjugate, and do the same with the denominator:

\left(\sqrt{2x+3}-\sqrt{3x}\right)\left(\sqrt{2x+3}+\sqrt{3x}\right)=\left(\sqrt{2x+3}\right)^2-\left(\sqrt{3x}\right)^2=-(x-3)

so that in the limit, we have

\displaystyle\lim_{x\to3}\frac{-(x-3)}{(x^2-3x)\left(\sqrt{2x+3}+\sqrt{3x}\right)}

Factorize the first term in the denominator as

x^2-3x=x(x-3)

The x-3 terms cancel, leaving you with

\displaystyle\lim_{x\to3}\frac{-1}{x\left(\sqrt{2x+3}+\sqrt{3x}\right)}

and the limand is continuous at x=3, so we can substitute it to find the limit has a value of -1/18.

7 0
3 years ago
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horsena [70]

Answer:

Can u show the whole question

8 0
2 years ago
Is this correct pls help
Lyrx [107]

Answer:

Volume of round cake pan = volume of cylinder

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As 108 > 76.97, the rectangular cake pan has a larger volume

4 0
2 years ago
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