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igor_vitrenko [27]
2 years ago
13

3000=30x+36y. Solve for x and y.

Mathematics
1 answer:
Lerok [7]2 years ago
8 0

Answer:x= 2(50-3y)/5

Step-by-step explanation:

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I will give brainiest to whoever answers correctly !!
Kisachek [45]
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2 years ago
3.) Simplify the expression by combining like
Evgesh-ka [11]

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3 0
2 years ago
The price of a mobile phone decreased from $375 to $250 during a clearance sale. What is the percent of decrease in the price?
Tresset [83]
So decrease is 375-250=75

so convert 75/375 to percent
percent means part out of 100 so 
x/100=x%

divide 125/375=0.3333/1

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4 0
3 years ago
Write the standard form of the equation of the circle with the given characteristics.
SOVA2 [1]
<h2><u>Circle Equations</u></h2>

<h3>Write the standard form of the equation of the circle with the given characteristics.</h3><h3>Center: (0, 0); Radius: 2</h3>

To determine the equation of a circle, use the standard form of a circle (x - h)² + (y - k)² = r² where,

  • <u>(h, k)</u> is the center; and
  • <u>r</u> is the radius

Substitute the values of the center and radius to the standard form.

<u>Given:</u>

<u>(0, 0)</u> - <u>center</u>

<u>2</u> - <u>radius</u>

  • (x - h)² + (y - k)² = 2²
  • (x - 0)² + (y - 0)² = 4
  • x² + y² = 4

<u>Answer:</u>

  • The equation of the circle is <u>x² + y² = 4</u>.

Wxndy~~

7 0
2 years ago
Someone help me with my math homework pleaseee. Find the volumes of the pyramids and the height is 7cm for the first one.
gregori [183]
\bf \textit{volume of a pyramid}\\\\&#10;V=\cfrac{1}{3}Bh\qquad &#10;\begin{cases}&#10;B=\textit{area of the base}\\&#10;h=height&#10;\end{cases}

now, the first one, on the far-left.... can't see the height.. but I gather you do, now as far as its Base area, well, the bottom is just a 12x12 square, so the area of its base is just 12*12


now, the middle pyramid, has a height of 6, the base is also a square, 8x8, so the Base area is just 8*8

now the last one on the far-right

has a height of 8, the Base is a Hexagon, with sides of 6

\bf \textit{area of a regular polygon}\\\\&#10;A=\cfrac{1}{4}ns^2cot\left( \frac{180}{n} \right)\qquad &#10;\begin{cases}&#10;n=\textit{number of sides}\\&#10;s=\textit{length of one side}\\&#10;\frac{180}{n}=\textit{angle in degrees}\\&#10;----------\\&#10;n=6\\&#10;s=6&#10;\end{cases}\\\\\\ A=\cfrac{1}{4}\cdot 6\cdot 6^2\cdot cot\left( \frac{180}{6} \right)
5 0
2 years ago
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