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kenny6666 [7]
3 years ago
5

TSU is shown

Mathematics
1 answer:
jekas [21]3 years ago
6 0
TS=8square root of 3
SU=24
TU=16 square root of 3

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A regular polygon
olga55 [171]

Answer:

3.6° = ext. angle

Step-by-step explanation:

For the shape to be a regular polygon, all interior angles must be congruent.

We can find every interior angle by using the formula:

\frac{180(n-2)}{n}

\frac{180(100-2)}{100}

\frac{180(98)}{100}

\frac{17640}{100}

176.4 = each int. angle

We also know that int. and ext. angles are supplementary!

180 = 176.4 - x

3.6° = ext. angle

5 0
3 years ago
A box of cereal weighs 12 ounces and costs $2.16. What is the price of the cereal per ounce?
patriot [66]
5.5555 is the answer
4 0
2 years ago
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Which statement best describes f(x)=-2√x-7+1
olga nikolaevna [1]

it is under mapping and function

f of x is equal to -2 root x-7 plus one

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3 years ago
PLEASE HELP DUE TODAY !
igor_vitrenko [27]

Answer:

let x be chips and y be sodas

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4x + 8y = 7------> second hint

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5 0
3 years ago
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How many pounds of a 15% copper alloy must be mixed with 700lb of a 30% copper alloy to maybe a 25.5% copper alloy
lisov135 [29]

Answer:

300\; \rm lb.

Step-by-step explanation:

Let x represent the mass (in pounds) of that 15\% copper alloy required, such that the final mixture would contain 25.5\% copper by mass.

Consider: if x pounds of that 15\% copper alloy is mixed with 700 pounds that 30\% copper alloy, what would be the mass of copper in the mixture?

  • Mass of copper in x pounds of that 15\% copper alloy: (0.15\, x)\; \rm lb.
  • Mass of copper in 700 pounds of that 30\% copper alloy: 700 \times 0.30 = 210\; \rm lb.

Therefore, the mixture would contain (210 + 0.15\, x) \; \rm lb of copper.

The mass of that mixture would be (700 + x)\; \rm lb. The mass fraction of copper in that mixture would be:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\%.

This ratio is supposed to be equal to 25.5\%. These two pieces of equations combine to give an equation about x:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\% = 25.5\%.

\displaystyle \frac{210 + 0.15\, x}{700 + x} = 0.255.

Simplify and solve for x:

210 + 0.15\, x= 0.255\, (700 + x).

(0.255 - 0.15)\, x= 210 - 0.255 \times 700.

\displaystyle x = \frac{210 - 0.255 \times 700}{0.255 - 0.15} = 300.

Therefore, 300\; \rm lb of that 15\% alloy would be required.

4 0
3 years ago
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