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Alexeev081 [22]
3 years ago
14

-9=-4/9x-13 algebra hw I'm confused on how to do it

Mathematics
1 answer:
Maurinko [17]3 years ago
4 0
Add 13 to both sides of equal sign and then divide by reciprocal of 4/9
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I'm not sure what this means, please help me I'll give you 10 points ;)
Karo-lina-s [1.5K]

<u>Answer:</u>

So all the possible solutions are:

  • \bold{523,~400,~379~and,~378.6}

<u />

<u>Solution Steps:</u>

<em>First you need to solve the real inequality to understand how to find the rest of the possible equations. </em>

<u>Add 78 to both sides:</u>

  • <u />78+78= Cancels Out
  • 300+78=378

<em>So now we know the real answer, but it ask for all possible answers. </em>

(Means anything larger than 300 when you plug it into x - 78 > 300.)

<u>Numbers that are greater than 378:</u>

  1.)  377-78=299 > 300  (False)

  2.)  523-78=445>300  (True)

  3.)  0-78=-78>300  (False)

  4.)  -62-78=-140>300  (False)

  5.)  400-78=322>300  (True)

  6.)  379-78=301>300  (True)

  7.)  222-78=144>300  (False)

  8.)  378.6-78=300.6>300 (True)

______________________________

 \bold{Hope~this~helps!}\\\bold{If~you~need~help~with~anything~else,~feel~free~to~ask!}\\\\\bold{~~~~~-TotallyNotTrillex}

5 0
3 years ago
Find the area of each figure.
grandymaker [24]

Answer: if thats a ten then its

Step-by-step explanation:

rectangle is LxW=  70

triangle is A=h(b)/b = 17.5

together it would be 87.5

i think lol

8 0
3 years ago
I have a math question I need help answering.
aniked [119]
The answer should be $21,840.50

Explanation:

20000*1.045=a (first year)
a*1.045=b (second year)
b = 21840.5
6 0
3 years ago
Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discha
mr Goodwill [35]

Answer:

Step-by-step explanation:

Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discharging hot water into the river is given by

T(x) = 160-0.05x^2

a. [0, 10]

For x = 0

T(0) = 160 - 0.05 × 0^2

T(0) = 160

For x = 10

T(10) = 160 - 0.05 × 10^2

T(10) = 160 - 5 = 155

The average temperature

= (160 + 155)/2 = 157.5

b. [10, 40]

For x = 10

T(10) = 160 - 0.05 × 10^2

T(10) = 160 - 5 = 155

For x = 40

T(10) = 160 - 0.05 × 40^2

T(10) = 160 - 80 = 80

The average temperature

= (80 + 155)/2 = 117.5

c. [0, 40]

For x = 0

T(0) = 160 - 0.05 × 0^2

T(0) = 160

For x = 40

T(10) = 160 - 0.05 × 40^2

T(10) = 160 - 80 = 80

The average temperature

= (160 + 80)/2 = 120

6 0
3 years ago
I have a lot of algebra problems. Someone help me even with this one please!
inysia [295]

Answer:

8 in

I have that is the correct answer

but iam not confident

3 0
3 years ago
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