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Paul [167]
3 years ago
5

How to simply the radical

Mathematics
1 answer:
zlopas [31]3 years ago
3 0
6√2 + 4 √32
try to get the same number "2" remaining under the root sign by finding what # times 2 gives 32
6√2 + 4√2×√16
since 16 is a perfect Square √16 gives 4
therefore.
6√2 + 4 ×4 √2
6√2 +16√2
these are Like terms so add them
which would give u
22√2
hoped i helped, feel free to ask any question XOX
You might be interested in
A radio transmission tower is feet tall. How long should a guy wire be if it is to be attached feet from the top and is tomake a
Bond [772]

Answer:

<em> Length of guy wire is 590.6 ft</em>

Step-by-step explanation:

The complete question is

A radio transmission tower is  210  feet tall. How long should a guy wire be if it is to be attached 8 feet from the top and is to make an angle of  20°  with the ground. Give your answer to the nearest tenth of a foot.

height of tower = 210 ft

8 ft fro the top leaves 210 - 8 = 202 ft to the ground

This is an angle of elevation problem

the opposite is 202 ft

hypotenuse = ?

angle is  20°  

using sin ∅ = opp/hyp

sin  20°  = 202/hyp

0.342 = 202/hyp

hyp = 202/0.342 = <em>590.6 ft   this is the length of the guy wire</em>

3 0
2 years ago
Posting a screenshot. please help.
Alinara [238K]
X^2 + 10x + 11 = 0
x^2 + 10x = -11
x^2 + 10x + 25 = -11 + 25
(x + 5)^2 = 14 <===
x + 5 = (+-) sqrt 14
x = -5 (+-) 3.74

x = -5 + 3.74 = -1.26 <== solution
x = -5 - 3.74 = - 8.74 <== solution


7 0
3 years ago
Suppose that each time Giannis Antetokounmpo shoots a free throw, he has a 3/4 probability of success. If Giannis shoots three f
GREYUIT [131]

Answer:

84.38% probability that he succeeds on at least two of them

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either Giannis makes it, or he does not. The free throws are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

He has a 3/4 probability of success.

This means that p = \frac{3}{4} = 0.75

Giannis shoots three free throws

This means that n = 3

What is the probability that he succeeds on at least two of them

P(X \geq 2) = P(X = 2) + P(X = 3)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{3,2}.(0.75)^{2}.(0.25)^{1} = 0.4219

P(X = 3) = C_{3,3}.(0.75)^{3}.(0.25)^{0} = 0.4219

P(X \geq 2) = P(X = 2) + P(X = 3) = 0.4219 + 0.4219 = 0.8438

84.38% probability that he succeeds on at least two of them

6 0
2 years ago
A spinner has 4 equal-sized sections labeled A, B, C, and D. It is spun and a fair coin is tossed. What is the probability of sp
allsm [11]
Probability of spinning a C =1/4
Probability of flipping heads = 1/2
1/4*1/2 = 1/8
8 0
2 years ago
Save me the headache
maxonik [38]

(9\sin2x+9\cos2x)^2=81

Taking the square root of both sides gives two possible cases,

9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1

or

9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1

Recall that

\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

If \alpha=2x and \beta=\dfrac\pi4, we have

\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}

so in the equations above, we can write

\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1

Then in the first case,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}

\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi

(where n is any integer)

\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi

\implies x=n\pi\text{ or }\dfrac\pi4+n\pi

and in the second,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}

\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi

\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi

\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi

Then the solutions that fall in the interval [0,2\pi) are

x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4

5 0
3 years ago
Read 2 more answers
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