All categories

2 years ago

Helpppppppppppppppppppp

Mathematics

1 answer:

Tju [1.3M]2 years ago

5 0

Answer:

the third option -24a^2b^12

Step-by-step explanation:

You might be interested in

JK and LM are straight lines. Find /NOK.<br> ( help me pls)<br>

VashaNatasha [74]

Answer:

$NOK = 58$

Step-by-step explanation:

Given

$JOM = 108$

$LON = 50$

Required

Find NOK

From the attached triangle, we have:

$LOK = JOM =108$ --- Corresponding angles

And

$LOK = LON + NOK$

Substitute for LOK and LON

$108= 50+ NOK$

Make NOK the subject

$NOK = 108 - 50$

$NOK = 58$

8 0

2 years ago

PLEASE HELP!!!!!!!! I. NEED. AN. ANSWER. LIKE. RIGHT. NOW!!!!!!!!!!

AVprozaik [17]

Answer:

butter brick and clay sink

oak and wheat float

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Can someone please help me with #12?

Genrish500 [490]

4x+12=90

4x=90-12

4x=8

x=8/4

x=2

5 0

2 years ago

18,000 divided by 25%

DanielleElmas [232]

Correct answer - 72000.

7 0

2 years ago

Read 2 more answers

Question regarding logarithms.

Eddi Din [679]

$5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}$

$\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$

$\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}$

6 0

2 years ago

Read 2 more answers