All categories

3 years ago

A pentagon has interior angles of 104°,

Mathematics

1 answer:

Elodia [21]3 years ago

7 0

Answer:

108

Step-by-step explanation:

Interior angles in a pentagon add up to 540 degrees. So we first need to find the total sum of all the known angles. We do this by calculating 104+108+112=324.

Now we need to find the total sum of the unknown angles, and we do this buy calculating 540-324=216.

The final step is to find the size of one of the 2 remaining angles, and since we know they are equal, we just have to divide by 2. So, 216/2=108. This is our final answer.

You might be interested in

Wastewater is filling barrels at the rate of 11 quarts per hour. The recycling facility picks up 120 full barrels on each trip,

givi [52]

120 barrels x 12 quarts per barrel = 1440 total quarts

1440 quarts / 11 quarts per hour = 130.91 hours to fill all the barrels

1 day = 24 hours

130.91 hours / 24 hours = 5.45 days = 5.5 days

5 0

3 years ago

NEED HELP ASAP PLEASE WITH #9 and #10!!

ICE Princess25 [194]

9) Because the total number of pencils is 180 and you will use them up in 30 days, the equation will have to equal 0 total pencils when 30 is substituted in for the time factor, or x. This already takes our choice 3 because it doesn’t meet this criteria.
The answer to how many pencils in 20 days could be answered by plugging in 20 for x. Choice 4 cannot work because it results in a negative number of pencils. Choices 1 and 2 use the same equation, so by plugging in 20 it is clear choice 2 is the correct answer.

10) A line parallel to the go en equation would have the same slope, -3, which means choices 1 and 4 are out. Plug in (-2,5) into both choices 2 and 3. Plugging -2 into x in choice 2 gives -5, and in choice 3 gives 5 for the y value. Therefore choice 3 is correct.

8 0

3 years ago

Lucas compared the prices of some books. He wrote down the prices of the books rounded to the nearest dollar as shown.6, 6, 7, 7

Anon25 [30]

Answer:

The third plot.

Step-by-step explanation:

Use the process of elimination:

The first 4 numbers are 6,6,7,7

The first plot shows 6,6,6,7,7,7 which is not the answer.

Next , the numbers are 9, 10, 10, 10, 10, 10, 10

THERE ARE 6 Tens, ONLY ANSWER CHOICE C HAS 6 TENS

THAT IS THE ANSWER!!!

<h3><u><em>Please rate this and please give brainliest. Thanks!!! </em></u></h3><h3><u><em>Appreciate it! : ) </em></u></h3><h3><u><em>And always, </em></u></h3><h3><u><em>SIMPLIFY BANANAS : ) Hope this helps!!! </em></u></h3>

3 0

3 years ago

Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))

NNADVOKAT [17]

I believe the given limit is

$\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)$

Let

$a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1$

Now rewrite the expression as a difference of cubes:

$a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}$

Then

$a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2$

The limit is then equivalent to

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}$

From each remaining cube root expression, remove the cubic terms:

$a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}$

$(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}$

$b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}$

Now that we see each term in the denominator has a factor of <em>x</em> ², we can eliminate it :

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}$

$=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}$

As <em>x</em> goes to infinity, each of the 1/<em>x</em> ⁿ terms converge to 0, leaving us with the overall limit,

$\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}$

8 0

3 years ago

PB_WINT FSA DIAG_MA_GRTA_682

8090 [49]

Really any option.

Explanation: A triangle can be obtuse of acute, either of these answers will make one of those.

4 0

3 years ago