All categories

3 years ago

Factor completely: A) 64x²-49y² B) 20x²-45y²

2 answers:

8 0

A) This is a difference of two squares. So your equation (a²x² - b²x²) where a and b can be square-rooted ends up being factored to (ax -by)(ax + by).

64x² - 49y² = (8x - 7y)(8x + 7y)

B) Factor out a GCF of 5 from both terms, then you have the same situation as equation A.

20x² - 45y² = 5(4x² - 9y²) = 5(2x - 3y)(2x + 3y)

tigry1 [53]3 years ago

3 0

$Use\\a^2-b^2=(a-b)(a+b)\\(ab)^n=a^nb^n\\(\sqrt{a})^2=a\ for\ a\geq0\\------------------------------\\A)\ 64x^2-49y^2=8^2x^2-7^2y^2=(8x)^2-(7y)^2=(8x-7y)(8x+7y)\\\\B)\ 20x^2-45y^2=(\sqrt{20})^2x^2-(\sqrt{45})^2y^2=(*)\\\\\sqrt{20}=\sqrt{4\cdot5}=\sqrt4\cdot\sqrt5=2\sqrt5\\\sqrt{45}=\sqrt{9\cdot5}=\sqrt9\cdot\sqrt5=3\sqrt5\\\\(*)=(2\sqrt5x)^2-(3\sqrt5y)^2=(2\sqrt5x-3\sqrt5y)(3\sqrt5x+3\sqrt5y)$

$or:\\\\20x^2-45y^2=5\cdot4x^2-5\cdot9y^2=5(4x^2-9y^2)=5[(2x)^2-(3y)^2]\\\\=5(2x-3y)(2x+3y)$

You might be interested in

Which of the following equals to 24 9/16 A 25.916 B 25.25 C 25.169 D 25.5625

Ulleksa [173]

C, because rounded up to the nearest tenth. 25.169 is closest to 24.

4 0

4 years ago

Which of the following functions is not odd?

guapka [62]

For odd function f(-x)= -f(x)

1) f(-x)= sin(-x) = -sin(x)=-f(x) sin is odd function
2) f(-x) = sin(-2x) = -sin 2x=-f(x)
3) f(-x)= (-x)³+1 = -x³ +1 (not an odd function)
To be an odd function, it should be like this:
if a function is (x³+1), to be odd f(-x) should be -(x³+1)=-x³-1
4) f(x)= x/(x²+1)
f(-x) = (-x)/((-x)²+1)= -x/(x²+1)=-f(x)
So, f(-x) gives $- \frac{x}{x^{2}+1}$ ,
that means that f(x)= x/(x²+1) is an odd function.
5)f(x) = ∛(2x)

f(-x) = ∛(2*(-x)=∛(2x*(-1)) = ∛(2x)*∛(-1)=- ∛(2x)= -f(x)

4 0

4 years ago

Read 2 more answers

The experimental probability that it will rain on any given day in Houston, Texas, is about 15%. Out of 365 days, about how many

abruzzese [7]

9. The experimental probability that it will rain on any given day in Houston Texas is about 15%. Thus, out of 365 days, 15% of that are the predicted days the it will rain in the Place. => 365 days is the total number of days in a year => 15% is the predicted days to rain => 365 * .15 = 54.75 Since this is a decimal number, let’s round this off to 55 days. Why 55 because the value of tenths is more than 5.

7 0

4 years ago

UCF believes that the average time someone spends in the gym is 56 minutes. The university statistician takes a random sample of

11111nata11111 [884]

Answer:

We conclude that the average time someone spends in the gym is different from 56 minutes.

Step-by-step explanation:

We are given that UCF believes that the average time someone spends in the gym is 56 minutes.

The university statistician takes a random sample of 32 gym goers and finds the average time of the sample was 50 minutes. Assume it is known the standard deviation of time all people spend in the gym is 8 minutes.

Let $\mu$ = population average time someone spends in the gym

SO, Null Hypothesis, $H_0$ : $\mu$ = 56 minutes {means that the average time someone spends in the gym is 56 minutes}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 56 minutes {means that the average time someone spends in the gym is different from 56 minutes}

The test statistics that will be used here is One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average time someone takes in the gym = 50 min

$\sigma$ = population standard deviation = 8 minutes

n = sample of gym goers = 32

So, test statistics = $\frac{50-56}{\frac{8}{\sqrt{32} } }$

= -4.243

Since in the question we are not given the level of significance so we assume it to b 5%. Now at 5% significance level, the z table gives critical value between -1.96 and 1.96 for two-tailed test. Since our test statistics does not lie within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the average time someone spends in the gym is different from 56 minutes.

4 0

3 years ago

On a cold day, hailstones fall with a velocity of 2i − 6k m s−1 . If a cyclist travels through the hail at 10i m s−1 , what is t

Paraphin [41]

Answer:

The velocity of the hail relative to the cyclist is $V_{hc}=-8\hat{i}-6\hat{k}$

The angle at which hailstones falling relative to the cyclist is $\theta = 36.86^\circ$

Step-by-step explanation:

Given : On a cold day, hailstones fall with a velocity of $2\hat{i}-6\hat{k}\text{ m/s}$ . If a cyclist travels through the hail at $10\hat{i} \text{ m/s}$ .