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lara31 [8.8K]
3 years ago
12

Factorize 2x^2-5/3xy-2y^2

Mathematics
2 answers:
steposvetlana [31]3 years ago
7 0

Answer:

y = 0 \: .x =  \frac{2}{3} y

Step-by-step explanation:

\frac{2 {x}^{2} - 5 }{3xy - 2 {y}^{2} }

3xy - 2 {y}^{2}  = 0

y \times (3x - 2y) = 0

y = 0 = 3x - 2y = 0

y = 0 = 3x = 2y

y = 0

x =  \frac{2}{3} y

y = 0. \: x =  \frac{2}{3} y

vladimir1956 [14]3 years ago
3 0

I hope it helps you

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I need your help broskis
Ilia_Sergeevich [38]

a very fast way to do it:

Trivially 8.8*5>10, so the power of 10 is 6+2+1 = 9.

Therefore, the answer is \boxed{4.4 \text{ x }10^9}

8 0
3 years ago
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A film distribution manager calculates that 4% of the films released are flops. If the manager is correct, what is the probabili
Anit [1.1K]

Answer:

9.34%

Step-by-step explanation:

p = 4%, or 0.04

n = Sample size = 667

u = Expected value = n * p = 667 * 0.04 = 26.68

SD = Standard deviation = \sqrt{np(1-p)}  =\sqrt{667*0.04*(1-0.04)} = 5.06

Now, the question is if the manager is correct, what is the probability that the proportion of flops in a sample of 667 released films would be greater than 5%?

This statement implies that the p-vlaue of Z when X = 5% * 667 = 33.35

Since,

Z = (X - u) / SD

We have;

Z = (33.35 - 26.68) / 5.06

Z = 1.32

From the Z-table, the p-value of 1.32 is 0.9066

1 - 0.9066 = 0.0934, or 9.34%

Therefore, the probability that the proportion of flops in a sample of 667 released films would be greater than 5% is 9.34%.

6 0
3 years ago
What's the ordered pair for point H in the graph above?
Bumek [7]

Answer:

C) (5,0)

Step-by-step explanation:

1. The correct answer is C) because when we move five units to right from the origin, and zero units up, we end up at point H!

6 0
3 years ago
5cosx -2sin(x/2) +7=0<br>Help me to find x step by step<br>pls​
miskamm [114]

Answer:

x = 180

Step-by-step explanation:

First, you need to know

1. Double-angle formula:

cos(2x) = cos^{2}x - sin^{2}x

2. Pythagorean identity:

cos^{2}x + sin^{2}x = 1

Back to your problem, replacing the variable by the above:

5cosx-sin\frac{x}{2}+7 = 0

5(cos^{2}\frac{x}{2}-sin^{2}\frac{x}{2}) - 2sin\frac{x}{2} + 7 = 0 By Double-angle formula

5(1 - 2sin^{2}\frac{x}{2}) - 2sin\frac{x}{2} + 7 = 0 By Pythagorean identity

Given y = \frac{x}{2}

5(1-2sin^{2}y) - 2siny + 7 = 0

10sin^{2}y+2siny-12=0

5sin^{2}y+siny-6=0

(5siny + 6)(siny - 1)=0, we know -1 < sinx < 1, for every x ∈ R

siny = 1, y =90

y = \frac{x}{2}

x = 180

8 0
3 years ago
Please help and explain how you did these problems :) help me with 14,16,18,20,22
igor_vitrenko [27]
You do the inverse operation for each one. so for #14, since 25g is multiplication you divide it by 25 to cancel it out, then you divide -175 by 25 which gives you -7
7 0
3 years ago
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