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3 years ago

Question 1 (1 point)

Mathematics

1 answer:

nasty-shy [4]3 years ago

8 0

Answer:

0 because both equations on either side will cancel out

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Does the system have no solution?

kvv77 [185]

Answer:

$\mathrm{A)\:}-6$

Step-by-step explanation:

Notice that in the second equation, the coefficient of $y$ is $2$ times greater than in the first. Therefore, by making the coefficient of $x$ in the second equation $2$ times greater than in the first, it's impossible to only isolate one variable.

Therefore, when $a=-3\cdot 2=\fbox{-6}$ , there are no solutions to this system.

6 0

3 years ago

Find the distance points (7,-2) (3,1)

grandymaker [24]

Answer:

square root of 25

Step-by-step explanation:

I used my ti-84 plus ce calculator to solve

>download DISTMID

>prgm

>DISTMID

>DISTENCE

>enter points

>answer

3 0

3 years ago

What is z over 5 - 7=2 and 1 3

dsp73

$\frac{z}{5-7}=2 \frac{1}{3} \\\\ -\frac{z}{2}=\frac{7}{3} \\\\ z=-\frac{2*7}{3} \\\\ \boxed{z=-\frac{14}{3}}$

5 0

3 years ago

Read 2 more answers

Limit xtens to 0 x^2logx^2 what is the ans of interminate forms?

GarryVolchara [31]

Rewrite the limit as

$\displaystyle\lim_{x\to0}x^2\log x^2=\lim_{x\to0}\frac{\log x^2}{\frac1{x^2}}$

Then both numerator and denominator approach infinity (with different signs, but that's not important). Applying L'Hopital's rule, we get

$\displaystyle\lim_{x\to0}\frac{\log x^2}{\frac1{x^2}}=\lim_{x\to0}\frac{\frac2x}{-\frac2{x^3}}=\lim_{x\to0}-x^2=\boxed{0}$

7 0

3 years ago

I really really need help!!!!!!

Svetllana [295]

ANSWER

$y = 5 \sin( \frac{6}{5} x - \pi)-4$

EXPLANATION

The given function is

$y = 5 \cos( \frac{3}{5} x - \pi) + 4$

The period of this function can be determined using the formula:

$T= \frac{2\pi}{ |b| }$

The period is

$T= \frac{2\pi}{ \frac{3}{5} } = \frac{10\pi}{3}$

Half the period of this function is:

$\frac{ \frac{10\pi}{3} }{2} = \frac{5\pi}{3}$

The function which has this period is

$y = 5 \sin( \frac{6}{5} x - \pi) -4$

The first choice is correct.

3 0

2 years ago