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3 years ago

Can someone please answer. There is a picture. There is only one problem. Thank you so much!!!

Mathematics

1 answer:

Lady_Fox [76]3 years ago

5 0

This is false, because again, a hyperbola could have /many/ different orientations and still technically be a hyperbola. It doesn't /in general/ have to be symmetric about the lines y = k or x = h, but it might be symmetric about, say, the line x + y = k + h, if it were rotated 45° relative to the "usual" equation. sorry for the long answer lol

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4. Find x<br> AB<br> (10x-51)<br> DC<br> (6x-11)

Volgvan

Answer:

51/10 and 11/6

Step-by-step explanation:

10x - 51 = 0

10x = 51

x = 51/10

6x - 11 = 0

6x = 11

x = 11/6

3 0

2 years ago

"Mia jogs 3 kilometers in 20 minutes. There are about 0.6 miles in a kilometer. What is Mia’s approximate speed in miles per min

Lapatulllka [165]

First we need to find Mia's speed (v) in km/min

v = distance travelled/ time
v = 3 km/20 minutes
v = 0.15 km /min

in order to convert the speed into mile/min, we need to use the conversion factor given in which for every kilometer, there is 0.6 miles.

v = 0.15 km/min *(0.6 miles/ km)
v =0.09 miles/min

therefore the speed in miles/min is 0.09 miles/min

4 0

4 years ago

Read 2 more answers

How do you write 9.11 x 10 4 in standard form?

vitfil [10]

Answer:

9400.11

Step-by-step explanation:

7 0

3 years ago

The proportion of students at a college who have GPA higher than 3.5 is 19%. a. You take repeated random samples of size 25 from

melomori [17]

Answer:

$\mu_{\hat{p}}=0.19$

$\sigma_{\hat{p}}=0.0785$

Step-by-step explanation:

We know that the mean and the standard error of the sampling distribution of the sample proportions will be :-

$\mu_{\hat{p}}=p$

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

, where p=population proportion and n= sample size.

Given : The proportion of students at a college who have GPA higher than 3.5 is 19%.

i.e. p= 19%=0.19

The for sample size n= 25

The mean and the standard error of the sampling distribution of the sample proportions will be :-

$\mu_{\hat{p}}=0.19$

$\sigma_{\hat{p}}=\sqrt{\dfrac{0.19(1-0.19)}{25}}\\\\=\sqrt{0.006156}=0.0784601809837\approx0.0785$

Hence , the mean and the standard error of the sampling distribution of the sample proportions :

$\mu_{\hat{p}}=0.19$

$\sigma_{\hat{p}}=0.0785$

8 0

3 years ago

If alpha and beta are the zeroes of the polynomial 6x2+x-2 find the value og alpha/beta + beta/alpha

castortr0y [4]

$6x^2+x-2=6x^2+4x-3x-2=2x(3x+2)-1(3x+2)\\\\=(3x+2)(2x-1)\\\\3x+2=0\to x=-\frac{2}{3}\\\\2x-1=0\to x=\frac{1}{2}\\\\\alpha=-\frac{2}{3};\ \beta=\frac{1}{2}\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{-\frac{2}{3}}{\frac{1}{2}}+\frac{\frac{1}{2}}{-\frac{2}{3}}=-\frac{2}{3}\cdot\frac{2}{1}-\frac{1}{2}\cdot\frac{3}{2}=-\frac{4}{3}-\frac{3}{4}\\\\=-\frac{16}{12}-\frac{9}{12}=-\frac{25}{12}=-2\frac{1}{12}$

$use\ Vieta's\ formula:\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{\alpha^2+2\alpha\beta+\beta^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2}{\alpha\beta}-2\\\\\alpha+\beta=\frac{-b}{a};\ \alpha\beta=\frac{c}{a}\\\\\frac{(\alpha+\beta)^2}{\alpha\beta}-2=\frac{\left(\frac{-b}{a}\right)^2}{\frac{c}{a}}-2=\frac{b^2}{a^2}\cdot\frac{a}{c}-2=\frac{b^2}{ac}-2$

$6x^2+x-2\\\\a=6;\ b=1;\ c=-2\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{1^2}{6\cdot(-2)}-2=\frac{1}{-12}-2=-2\frac{1}{12}$

7 0

3 years ago