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aleksandr82 [10.1K]
3 years ago
8

3x + 3 = x - 5 Solve for x

Mathematics
2 answers:
OLga [1]3 years ago
7 0
Answer:
-4
Explanation:
3(-4)+3=-4-5
^ ^
-12 -9
-12+3=-9
^
-9 =-9
Mars2501 [29]3 years ago
4 0

Answer:

x = -4

Step-by-step explanation:

plz vote brainliest :)

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G(x)=-x^2/4 +7<br> Over which interval does g have a negative average rate of change?
Sholpan [36]

Answer:

x = 49

Step-by-step explanation:

1- Substitute g ( x ) = 0

Reduce

g ( x ) = - x 2/4 + 7

2- Move the variable to the left

0 = - x 1/2 + 7

3- Simplify the equation

x 1/2 = 7

4- Simplify Evaluate

( x 1/2 ) ² = 7²

5- Check the solution

x= 49

6- Simplify

0= - 49 2/4 + 7

7- x = 49 is a solution

0 = 0  

4 0
3 years ago
Put in simplest form 8/15 × 5/6
algol13
8/15 x 5/6
•multiply across the numerator and denominator.

= 40/90
• then find a common factor that will go into both the numerator and denominator equally.

Common factor: 5

= 8/18
•there is another common factor, which is 2

Simplified:

=4/9
6 0
3 years ago
Can you please hurry I only have like 25 minutes left
Sholpan [36]

Answer:

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Find the sum of the first five terms using the geometric series formula for the sequence 1/2, -1,2,-4…
sammy [17]

The sum of the first five terms is 11/2

<h3>How to determine the sum?</h3>

The series is given as:

1/2, -1,2,-4…

Calculate the common ratio of the series using

r = -4/2

Evaluate

r = -2

The sum is then calculated as:

S_n = a\frac{1 - r^n}{1 - r}

This gives

S_n = \frac{1}{2} * \frac{(1 - (-2)^5}{1 + 2}

Evaluate

S_n = \frac{1}{2} * \frac{33)}{3}

Evaluate

S_n =\frac{11}{2}

Hence, the sum of the first five terms is 11/2

Read more about series at:

brainly.com/question/11346378

#SPJ1

7 0
1 year ago
In a herd of cattle, the ratio of the number
Neko [114]

Answer:

In a herd no. of Cattle are: x

The ratio of bulls to cows is 1:6

\frac{bulls}{cows}  =  \frac{1}{6}  

6 bulls = 1 cows

bulls \:  + cows = x

bulls + 6 \: bulls \:  = x

7bulls = x

7 0
3 years ago
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