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2 years ago

Write an equation that represents the line. use exact numbers.

Mathematics

1 answer:

nekit [7.7K]2 years ago

3 0

Answer The line two origins are 6 and 5.1

Step-by-step explanation:

Hope this helps ^^

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1-m<br> ___ =2/3 HELP PLEASE<br> 1+m

marusya05 [52]

$Domain:\\\\1+m\neq0\to m\neq-1\\\\\dfrac{1-m}{1+m}=\dfrac{2}{3}\qquad\text{cross multiply}\\\\3(1-m)=2(1+m)\qquad\text{use distributive property}\\\\(3)(1)+(3)(-m)=(2)(1)+(2)(m)\\\\3-3m=2+2m\qquad\text{subtract 3 from both sides}\\\\-3m=-1+2m\qquad\text{subtract 2m from both sides}\\\\-5m=-1\qquad\text{divide both sides by (-5)}\\\\m=\dfrac{1}{5}\neq-1\\\\Answer:\ \boxed{m=\dfrac{1}{5}}$

7 0

3 years ago

Arthur walked around a circular trail twice in a total of 4.5 hrs. He walked the circle at 4 mph the first time and at 5 mph the

Alex

Answer:

It’s 5 mph I would be tired if I had to walk 4 miles

Step-by-step explanation:

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2 years ago

You earn $8.50 per hour at your summer job right and solve any quality any quality that represents the number of hours you need

irina1246 [14]

He should work for 24 hours.

7 0

2 years ago

Conner earns 24$ working every 1.5 12 blank

RoseWind [281]

Answer: That makes no sense. Maybe did your copy and paste corrupt? Or mistyped the question? Get back to me and I’ll help!!

Step-by-step explanation:

5 0

1 year ago

CALC- limits<br> please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

3 0

2 years ago

Write an equation that represents the line. use exact numbers.​

Write an equation that represents the line. use exact numbers.