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Ganezh [65]
3 years ago
15

CAN SOMEONE PLEASE HELP!!!! determine the value of x. i’ll give you points if you answer!

Mathematics
1 answer:
salantis [7]3 years ago
7 0

Answer:

2.4

Step-by-step explanation:

Look up 30-60-90 right triangle and use that formula. Since the angle in the middle is 90 degrees, it should work. The side of x would be double of the given side 1.2... this is what I would do

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70+95 Factor the expression using the GCF <br><br> giving brain
GenaCL600 [577]

Answer:

70 + 95 = 5(14 + 19)

70 + 95 = 5(14 + 19)

gcf = 5

Step-by-step explanation:

70+95

70=2*5*7

95=5*19

70+95=(5*2*7)+(5*19)=(5*14)+(5*19)=5*(14+19)=5*(33)

5*33=165 and 70+95=165                                                                                                  Hope this helps! Please give me brainly!      

6 0
2 years ago
One of the factors of 6x3 − 864x is<br><br> A. 4<br> B. X^2<br> C. X + 12<br> D. X - 8
erastova [34]
The answer is C. X+12
8 0
2 years ago
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Solving proportion 10\8=n\10
Zanzabum
\frac { 10 }{ 8 } =\frac { n }{ 10 } \\ \\ 10\cdot \frac { 10 }{ 8 } =\frac { n }{ 10 } \cdot 10\\ \\ \frac { 100 }{ 8 } =n

\\ \\ n=\frac { 4\cdot 25 }{ 4\cdot 2 } \\ \\ \therefore \quad n=\frac { 25 }{ 2 }
6 0
2 years ago
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What is the equation for the line in slope intercept form PLEASE HELP​
aliina [53]

Answer:

y=-4x+5

Step-by-step explanation:

6 0
2 years ago
Which of the following operations is true regarding relative frequency distributions? Multiple choice question. No two classes c
pshichka [43]

Answer:

The relative frequency is found by dividing the class frequencies by the total number of observations

Step-by-step explanation:

Relative frequency measures how often a value appears relative to the sum of the total values.

An example of how relative frequency is calculated

Here are the scores and frequency of students in a maths test

Scores (classes)              Frequency                Relative frequency

0 - 20                                10                               10 / 50 = 0.2

21 - 40                               15                               15 / 50 = 0.3

41 - 60                               10                               10 / 50 = 0.2

61 - 80                                5                                 5 / 50  = 0.1

81 - 100                             <u> 10</u>                                10 / 50 = <u>0.2</u>

                                          50                                               1

From the above example, it can be seen that :

  1. two or more classes  can have the same relative frequency
  2. The relative frequency is found by dividing the class frequencies by the total number of observations.
  3. The sum of the relative frequencies must be equal to one
  4. The sum of the frequencies and not the relative frequencies is equal to the number of observations.

4 0
2 years ago
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