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dangina [55]
1 year ago
6

Find an angle between 0 degrees and 360 degrees and that is coterminal with 1260 degrees.

Mathematics
1 answer:
GarryVolchara [31]1 year ago
7 0

Answer:

<h2><u><em>The 180 degrees angle.</em></u></h2>

Step-by-step explanation:

We can start by dividing 1260 / 360 = 1080. Therefore, a 1080 degrees angle would be coterminal with 360.

Now we take 1260 - 1080, to see how many degrees we have left, this gives us: 180, therefore, the angle between 0 and 360 that is coterminal with 1260 degrees is the 180 degrees angle.

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Genuinely confused, how would you solve this? (will mark for brainliest!!)
Yanka [14]

Answer:

Length of the bold sector: 40.3ft

Area of the bold sector: 282.08

Step-by-step explanation:

Finding the sector length: Assuming that pi=3.14, the diameter is 87.92 ft. The ratio of the bold section to the whole circumference is 165 to 360. So you need to multiply the circumference by the ratio. (87.92 x 165)/360=40.3

Finding the sector area: Assuming that pi=3.14, the area of the circle is 615.44 sq. ft. The ratio of the sector area to the area of the whole circle is 165:360. (615.44 x 165)/360=282.08, rounded to the nearest 100th.

3 0
3 years ago
Evaluate -4a + 10 when a = -2
stiks02 [169]

Answer:

<u>18</u>

Step-by-step explanation:

<em>replace a with -2</em>

-4(-2) + 10

<em>multiply the -4 and the -2, the negatives cancel out and leave you with 8</em>

8 +10

<em>just add</em>

18

4 0
3 years ago
Martin has a vegetable garden that is 8 feet long and 5 feet wide. Next year, Martin wants to increase both the length and width
KatRina [158]

Answer:

9.2 feet long and 5.75 feet wide

4 0
3 years ago
21.12 is what percent of 82.5 ?
Tanya [424]
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5 0
3 years ago
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
2 years ago
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