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nikklg [1K]
3 years ago
11

Help please I do t have lot of time

Mathematics
1 answer:
lys-0071 [83]3 years ago
4 0
It’s b my good friend
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- {(-2, 6), (2.0), (3,6), (4, -1), (5,3)} is it a function​
nikitadnepr [17]
Yes because none of the x’s are the same
3 0
3 years ago
What’s the answer for 7+25•4
kotykmax [81]
Use Order of Operations!! Remember PEMDAS!
P: Parentheses first
E: Exponents (ie Powers and Square Roots, etc.)
MD: Multiplication and Division (left-to-right)
AS: Addition and Subtraction (left-to-right)

7+25•4 (multiply 25 by 4 first)
7+100 (then add 7 to 100)
107

So your final answer would be 107

Hope this helped and I hope you have an awesome day!! :D
3 0
3 years ago
Read 2 more answers
PLEASE HELP I NEED THIS TO BE DONE
yanalaym [24]
To find the mean you add all the #'s and divide by how many there are. 4+5+6+7 +8= 30 /5 is 6. Tara is right. 
5 0
3 years ago
Can anyone solve this triangle?
kirill [66]

Answer:

Step-by-step explanation:

a^2+6^2 = 23^2

a^2 = 23^2 - 6^2

a^2 = 529 - 36

a^2 = 493

a = √493

a = 22,20

7 0
3 years ago
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
3 years ago
Read 2 more answers
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