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ratelena [41]
4 years ago
14

Help I need helppp. plz help bros​

Mathematics
1 answer:
ladessa [460]4 years ago
7 0

Answer:

x= 7

Step-by-step explanation:

5x - 2 + x = 9 + 3x + 10

6x - 2 = 19 + 3x

-3x                  -3x

3x - 2 =  19

    +2     +2

3x = 21

x = 7

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Them sum of two numbers is 17. One number is 3 less than 2/3 of the numbers. What is the lesser number?
Flauer [41]

Answer:

The smaller number is 5

Step-by-step explanation:

Let the smaller number be x

the other number is (17 -x) (we know they add up to 17)

2/3 of the bigger number is written as 2/3 (17 - x)

The smaller number is 3 less than that. (so subtract 3 to get x)

x= 2/3 (17 -x) - 3 <-- We have an equation, solve for x

Check 2/3 x 12 - 3 = 8 - 3 = 5

4 0
3 years ago
Let f(x) = x2- 3x - 7. Find f(-3)
Alexxx [7]
F(x) = x^2 - 3x - 7
f(-3) = (-3)^ 2 - 3(-3) - 7
f(-3) = 9 + 9 - 7
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6 0
3 years ago
The graph for the equation y = 2 x minus 2 is shown below. On a coordinate plane, a line goes through points (0, negative 2) and
weqwewe [10]

Answer:

i think it is b correct me if i am rong

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
I really need help with this pls.
prisoha [69]
H= V / pi r^2

and 

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3 0
3 years ago
Express the null hypothesis and the alternative hypothesis in symbolic form. The owner of a football team claims that the averag
Brilliant_brown [7]

Answer:

H_{0}: \mu \leq 63500\\H_A: \mu > 63500

Step-by-step explanation:

We are given the following in the question:

The owner of a football team claims that the average attendance at games is over 63,500.

He wants to justify that the team needs to be moved to a larger stadium outside the city.

If the attendance is larger than 63,500 the team would be moved to a larger stadium and if it is less than or equal to 63,500 that it would not.

Thus, the null and alternate hypothesis will be designed as:

H_{0}: \mu \leq 63500\\H_A: \mu > 63500

The null hypothesis says that the average attenders is equal to or less than 63,500 and alternate supports the claim that the attenders average is greater than 63,500.

4 0
3 years ago
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