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monitta
3 years ago
6

At Jefferson Middle School, 3/5 of the students play a sport. Of those students, 1/3 play soccer. What equation can be used to f

ind what fraction of the students play soccer
A 1/3 - 3/5
B 1/3 x 3/5
C 1/3 + 3/5
D 1/3 ÷ 3/5
Mathematics
1 answer:
yaroslaw [1]3 years ago
7 0

Answer:

a

Step-by-step explanation:

I think that right so let me know if It is right

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Suppose there is a 1.7°F drop in temperature for every thousand feet that an
kotegsom [21]

Answer: 61.2?

Step-by-step explanation:

5 0
3 years ago
Which is the value of this expression when X=-2
e-lub [12.9K]
A) x^2 +x -30 = 0 is factored by looking for two factors of 30 that differ by 1. We know ...  30 = 1*30 = 2*15 = 3*10 = 5*6The last two factors differ by 1, so we can factor the trinomial as  (x +6)(x -5) = 0
b) The solutions are found by finding values of x that make these factors zero. The only way the product will be zero is if one or more of the factors is zero.  x + 6 = 0  x = -6 . . . . . subtract 6
  x - 5 = 0  x = 5 . . . . . add 5
The solutions are x = -6 or x = 5These are the values of x that will satisfy the equation (make it true). What they mean depends on the meaning of the variable and the situation the equation is a model of.
6 0
3 years ago
I really need help with this Triangle problem. ​
julia-pushkina [17]

Answer:

x = 10

y = 20

Step-by-step explanation:

\tan \: 30 \degree =  \frac{x}{10 \sqrt{3} }  \\  \\  \frac{1}{ \sqrt{3} } =  \frac{x}{10 \sqrt{3} }  \\  \\ \sqrt{3} x = 10 \sqrt{3}  \\  \\ x =  \frac{10 \sqrt{3} }{ \sqrt{3} }  \\  \\ x = 10 \\  \\  \cos 30 \degree =  \frac{10 \sqrt{3} }{y}  \\  \\  \frac{ \sqrt{3} }{2}  =  \frac{10 \sqrt{3} }{y}  \\  \\ y =  10 \sqrt{3}  \times  \frac{2}{ \sqrt{3} }  \\  \\ y = 10 \times 2 \\  \\ y = 20

4 0
3 years ago
a $1,000 savings account earns 2.3% interest compounded monthly. how long will it take to double the original investment?
Rudik [331]
It will take a total of 44 months
3 0
3 years ago
For the functions f(x) = x^2+ 8x + 2 and g(x) = -5+9, find (f•g)(x)and (f.g)(-1).
Ne4ueva [31]

Answer:  (f·g)(x) = -5x³ - 31x² + 62x + 18

               (f·g)(-1) = -70

(fog)(x) = 25x² - 130x + 155

(fog)(-1) = 310

<u>Step-by-step explanation:</u>

f(x) = x² + 8x + 2          g(x) = -5x + 9

(f·g)(x) = (x² + 8x + 2)(-5x + 9)

          = -5x³ + 9x²

                    - 40x² + 72x

            <u>                    - 10x + 18</u>

          = -5x³ - 31x² + 62x + 18

(f·g)(-1)= -5(-1)³ - 31(-1)² + 62(-1) + 18

         =  -5(-1)  - 31(1)    - 62      + 18

         =    5     -   31      - 62      + 18

         =  -70

****************************************************************************************

(fog)(x) = (-5x + 9)² + 8(-5x + 9) + 2

           = 25x² - 90x + 81

                       - 40x + 72

              <u>                   +   2</u>

           = 25x² - 130x + 155

(fog)(-1) = 25(-1)² - 130(-1) + 155

            =   25    +  130    + 155

            = 310

<em>It wasn't clear if you wanted multiplication or composition so I solved both.</em>

6 0
3 years ago
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