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elena-14-01-66 [18.8K]
3 years ago
12

Which of the following is a trinomial with a constant term?

Mathematics
2 answers:
Alex3 years ago
8 0

Answer:

A

Step-by-step explanation:

only the first one has a number with no variables near

Pepsi [2]3 years ago
4 0

Answer:

is A

Step-by-step explanation:

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11 m
liubo4ka [24]

first off let's notice that the height is 11 meters and the volume of the cone is 103.62 cubic centimeters, so let's first convert the height to the corresponding unit for the volume, well 1 meters is 100 cm, so 11 m is 1100 cm.

\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ V=\stackrel{cm^3}{103.62}\\ h=\stackrel{cm}{1100} \end{cases}\implies 103.62=\cfrac{\pi r^2 (1100)}{3} \\\\\\ 3(103.62)=1100\pi r^2\implies \cfrac{3(103.62)}{1100\pi }=r^2 \\\\\\ \sqrt{\cfrac{3(103.62)}{1100\pi }}=r\implies \stackrel{cm}{0.00510199305952} \approx r

6 0
2 years ago
Evaluate the expression. Enter your answer in the box. 2³ =
masya89 [10]
2^3 = 2 x 2 x 2 = 8

your answer is 8

hope this helps
3 0
3 years ago
Read 2 more answers
Solve for y.<br> -1 = 8+3y<br> Simplify you answer as much as possible.
OleMash [197]

Answer:

-3

Step-by-step explanation:

8+3y = -1\\3y = -9\\y = -3

8 0
3 years ago
Read 2 more answers
What's the prime factorization of 78​
Masja [62]

Answer:

2,3 &13

Step-by-step explanation:

78=2times39=3times13

therefore the prime factors of 78=2,3,13

4 0
2 years ago
Evaluate: <img src="https://tex.z-dn.net/?f=%5Csqrt%5B-3%5D%7B27%7D" id="TexFormula1" title="\sqrt[-3]{27}" alt="\sqrt[-3]{27}"
GenaCL600 [577]

Answer:

1/3 is the simplified form for given expression.

Step-by-step explanation:

Given that:

=\sqrt[-3]{27}

By simplifying:

Radical sign will be removed as follows:

= 27^{-1/3}

For removing the "-" sign from power, base will be inverted:

= (1/27) ^ {1/3}

27 can also be written as 3 * 3 * 3 = 3^3

So,

= (1/3^3)^{1/3}

= 1^{1/3} / 3^{3* 1/3}

By simplifying we get:

= 1/3

i hope it will help you!

6 0
3 years ago
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