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3 years ago

How to write in a single power?

Mathematics

1 answer:

nekit [7.7K]3 years ago

7 0

Write what??
<em>n single power??</em>

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Is there a series of rigid transformations that could map ΔQRS to ΔABC? If so, which transformations could be used?

faltersainse [42]

Answer;

D Yes, ΔQRS can be translated so that Q is mapped to A and then reflected across the line containing QS.

Explanation;

-A rigid transformation is a transformation of a plane that preserves length. It is also called isometry. Reflections, translations, rotations, and combinations of these three transformations are examples of rigid transformations.

3 0

3 years ago

Read 2 more answers

What is the value of the expression 8a + 16c if a = -5 and C = -1

marysya [2.9K]

Answer:

-4 is the answer

Step-by-step explanation:

Mark brainliest please if my answer is correct

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3 years ago

A box contains 6 red marbles, 4 green marbles, 3 blue marbles, and 2 yellow marbles. If Greg chooses one marble at random 300 ti

dusya [7]

Answer:

$60$ times

Step-by-step explanation:

There are $6+4+3+2=15$ marbles in total. Out of those, $3$ are blue, so the probability of pulling a blue marble is $\frac{3}{15}=\frac{1}{5}$ . However, this is only the probability of pulling a blue marble when $1$ marble is pulled at random. Therefore, we can expect a blue marble to be pulled $\frac{1}{5}*300=60$ times if $1$ marble is pulled $300$ times at random. Hope this helps!

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3 years ago

Solve for x ????????????

Novosadov [1.4K]

Answer:

x = 8°

Step-by-step explanation:

since the angles are vertical angles, they are equal

3 + 14x = 115

14x = 112

x = 8

4 0

3 years ago

A box holds 8 red and 4 blue beads. Three beads are taken from the box and not replaced. Determine: The probability that all thr

zubka84 [21]

Answer: $\dfrac{14}{55}, \dfrac{13}{55}$

Step-by-step explanation:

Given

There are 8 red and 4 blue beads

Three beads were taken from the box

(i)No of ways of choosing three red beads out of 12 beads is $^8C_3$

Total no of ways $^{12}C_3$

Probability is

$P=\dfrac{^8C_3}{^{12}C_3}=\dfrac{8\times 7\times 6}{12\times 11\times 10}=\dfrac{2\times 7}{11\times 5}\\P=\dfrac{14}{55}$

(ii)atleast two beads refers to minimum two beads

There can be two possibilities (2B,1 R), (3B, 0R)

$\Rightarrow P=\dfrac{^4C_2\times ^8C_1}{^{12}C_3}+\dfrac{^4C_3}{^{12}C_3}\\\\\Rightarrow P=\dfrac{6\times 8}{220}+\dfrac{4}{220}=\dfrac{52}{220}\\\\\Rightarrow P=\dfrac{13}{55}$

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3 years ago