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3 years ago

There are two nets that contain circles. What statement is true about both nets?

Mathematics

1 answer:

dezoksy [38]3 years ago

5 0

Answer:

there is one circle

Step-by-step explanation:

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A conjecture and the paragraph proof used to prove the conjecture are shown.

amm1812

Answer:

Here is the correct answer. I just took the test.

Step-by-step explanation:

8 0

4 years ago

What is 3/10 + 29/100 = ?<br><br> What is the answer?

Westkost [7]

The answer is 59/100. First, you have to change 3/10 to 30/100 and then add that by 29/100 to get your answer

3 0

3 years ago

Read 2 more answers

Jacob the dog eats dry food that contains 324kcal/cup. Jacob eats 3.5 cups a day. The food contains 3.6 g of fat in 100 kcals of

natita [175]

Answer:

His daily fat intake is 40.824 g .

His daily Vitamin D intake is 290.304 IU .

Step-by-step explanation:

1 cup contains 324 kcal

Jacob eats 3.5 cups a day.

So, total calories in Jacob food = $324 \times 3.5$

= $1134 kcal$

Amount of fat in 100 kcals = 3.6g

Amount of fat in 1 kcal = $\frac{3.6}{100}$

Amount of fat in 1134 kcal = $\frac{3.6}{100} \times 1134$

= $40.824g$

So, his daily fat intake is 40.824 g .

Amount of vitamin D in 100 kcals = 25.6 IU

Amount of vitamin D in 1 kcal = $\frac{25.6}{100}$

Amount of vitamin D in 1134 kcal = $\frac{25.6}{100} \times 1134$

= $290.304$

So, His daily Vitamin D intake is 290.304 IU .

7 0

3 years ago

Simplify 81p6q2÷3p2q5 HELPPP

goldfiish [28.3K]

Answer:

$\displaystyle \frac{81\, p^{6}\, q^{2}}{3\, p^{2} \, q^{5}} = \frac{27\, p^{4}}{q^{3}}$ .

Step-by-step explanation:

Make use of the fact that for any $x \ne 0$ and integer $n$ :

$\displaystyle \frac{1}{x^{n}} = x^{-n}$ .

For example, in this question:

$\displaystyle \frac{1}{p^{2}} = p^{-2}$ .

$\displaystyle \frac{1}{q^{5}} = q^{-5}$ .

Thus, the original expression would be equivalent to:

$\begin{aligned}& \frac{81\, p^{6}\, q^{2}}{3\, p^{2} \, q^{5}} \\ =\; & \frac{81}{3}\, (p^{6}\, p^{-2})\, (q^{2}\, q^{-5}) \\ =\; & 27\, (p^{6}\, p^{-2})\, (q^{2}\, q^{-5})\end{aligned}$ .

Also make use of the fact that for any $x \ne 0$ , integer $m$ , and integer $n$ :

$x^{m}\, x^{n} = x^{m + n}$ .

Thus:

$\begin{aligned}& 27\, (p^{6}\, p^{-2})\, (q^{2}\, q^{-5}) \\=\; & 27\, p^{6 + (-2)}\, q^{2 + (-5)} \\ =\; & 27\, p^{4}\, q^{-3} \\ =\; & \frac{27\, p^{4}}{q^{3}}\end{aligned}$ .