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3 years ago

Which triangles are similar to AABC? B 10 С 10 А

Mathematics

1 answer:

Talja [164]3 years ago

6 0

Khan academy I have it too. I will answer your question since you like khan academy
The answer is really C the reason I know is because I too have done this before

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If a+b+c=16 and a^2+b^2+c^2=90 then find the value a^3+b^3+c^3-3abc

grin007 [14]

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\\(16)^2=90+2(ab+bc+ca)\\256-90=2(ab+bc+ca)\\166/2=ab+bc+ca\\83=ab+bc+ca\\a^3-b^3-c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))\\16(90-(83))\\16(7)\\=112$

3 0

3 years ago

D(x)=(x^2-12x+20)/(3x)

krok68 [10]

Answers:

Vertical asymptote: x = 0

Horizontal asymptote: None

Slant asymptote: (1/3)x - 4

Explanation:

d(x) = $\frac{x^{2}-12x+20}{3x}$

= $\frac{(x-2)(x - 10)}{3x}$

Discontinuities: (terms that cancel out from numerator and denominator):

Nothing cancels so there are NO discontinuities.

Vertical asymptote (denominator cannot equal zero):

3x ≠ 0

÷3 ÷3

x ≠ 0

So asymptote is to be drawn at x = 0

Horizontal asymptote (evaluate degree of numerator and denominator):

degree of numerator (2) > degree of denominator (1)

so there is NO horizontal asymptote but slant (oblique) must be calculated.

Slant (Oblique) Asymptote (divide numerator by denominator):

(1/3)x - 4
3x) x² - 12x + 20
x²
-12x
-12x
20 (stop! because there is no "x")

So, slant asymptote is to be drawn at (1/3)x - 4

6 0

3 years ago

I don't know who needs to hear this but I wanted to hopefully reach out to someone.

lora16 [44]

Answer:

oh my- this actually made me happy-

Step-by-step explanation:

8 0

3 years ago

Last year, 50% of MNM. Inc. employees were female. It is believed that there has been a reduction in the percentage of females i

Serhud [2]

Answer:

We conclude that there has been a significant reduction in the proportion of females.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 400

p = 50% = 0.5

Alpha, α = 0.05

Number of women, x = 118

First, we design the null and the alternate hypothesis

$H_{0}: p = 0.50\\H_A: p < 0.50$

This is a one-tailed test.

Formula:

$\hat{p} = \dfrac{x}{n} = \dfrac{118}{400} = 0.45$

$z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Putting the values, we get,

$z = \displaystyle\frac{0.45-0.50}{\sqrt{\frac{0.50(1-0.50)}{400}}} = -2$

Now, we calculate the critical value.

Now, $z_{critical} \text{ at 0.05 level of significance } = -1.645$

Since the calculated z-statistic is less than the critical value, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, there has been a significant reduction in the proportion of females.

3 0

3 years ago

PLEASE I NEED HELP PLEASE

mr_godi [17]

Answer:

none

Step-by-step explanation:

there's only ten cubes

3 0

2 years ago