All categories

3 years ago

What is the answer to 10m + 7=-7m +-11

Mathematics

2 answers:

den301095 [7]3 years ago

5 0

Just simplify?
Than your answer should be $m=- \frac{18}{17}$

Harlamova29_29 [7]3 years ago

3 0

<span>Simplifying -10m + 7 = -7m + -11 Reorder the terms: 7 + -10m = -7m + -11 Reorder the terms: 7 + -10m = -11 + -7m Solving 7 + -10m = -11 + -7m Solving for variable 'm'. Move all terms containing m to the left, all other terms to the right. Add '7m' to each side of the equation. 7 + -10m + 7m = -11 + -7m + 7m</span>

You might be interested in

Aaron needs a total of $410 to buy a new bicycle. He has $35 saved. He earns $15 each week delivering newspapers. How many weeks

Ostrovityanka [42]

35+15x=410
410-35=375
375/15=25
He will need to work 25 weeks.

3 0

3 years ago

Given m|n, find the value of x.

yaroslaw [1]

X=12

Because 180-168= 12

7 0

3 years ago

A music store sells electric guitars in 3 colors and 2 sizes. How many different guitar can you buy

matrenka [14]

Answer:

the answer is 6

Step-by-step explanation:

3 times 2 equals 6

5 0

3 years ago

In triangle MPN, angle P is a right angle. If cos N= 3/5 which of the following statements is true

Bezzdna [24]

Answer: Sin N = $\frac{4}{5}$

Step-by-step explanation:

Since we have given that

In triangle MPN, ∠P=90°

$\cos N=\frac{3}{5}$

As we know the formula for "Trigonometric Ratios":

$\cos N=\frac{Base}{Hypotenuse}=\frac{3}{5}$

By Pythagorus theorem , we get

$H^2=B^2+P^2\\\\5^2=3^2+P^2\\\\25=9+P^2\\\\25-9=P^2\\\\16=P^2\\\\P=\sqrt{16}\\\\P=4\ units$

So, we get,

$\sin N=\frac{Perpendicular}{Hypotenuse}=\frac{4}{5}$

Hence, Sin N = $\frac{4}{5}$

3 0

3 years ago

Find the derivative of<br> <img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Cfrac%7B6%7D%7Bx%7D%20" id="TexFormula1" titl

aev [14]

$f'(x_0)=\lim\limits_{h\to0}\dfrac{f(x_0+h)-f(h)}{h}=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{6}{x};\ x_0=2\\\\subtitute\\\\f'(2)=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-\frac{6}{2}}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-3}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6-3x}{x}}{x-2}=\lim\limits_{x\to 2}\dfrac{6-3x}{x(x-2)}\\\\=\lim\limits_{x\to2}\dfrac{-3(x-2)}{x(x-2)}=\lim\limits_{x\to2}\dfrac{-3}{x}=-\dfrac{3}{2}=-1.5\\\\\\An swer:\boxed{f'(2)=-1.5}$

8 0

3 years ago