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AlekseyPX
2 years ago
9

Three-fifths of the members of the Spanish club are girls. There are a total of 30 girls in the Spanish club. Which statements c

an be used to solve for x, the total number of members in the Spanish club? Select three options. Okx X = 30 3 х 5 - 30 3 513 X = 5 of(x=30(2) x=(2 35 53 X= 30 3 O x = 50​

Mathematics
1 answer:
arsen [322]2 years ago
4 0
Divide by 3 then subtract however many you have to
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Solve the equation. Show all work.<br> -2 (4 - x) = 12x – 3<br> (I think this is 9th grade algebra)
Genrish500 [490]

Answer:

-1/2

Step-by-step explanation:

-2 (4 - x) = 12x – 3

-8+2x=12x-3

+8            +8

2x=12x+5

-12x  -12x

-10x=5

/-10  /-10

= -1/2

4 0
2 years ago
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What is the value of m, if the equation my2+2y−4=0 has exactly one root?
stich3 [128]

Answer:

m=\frac{-1}{4}

Step-by-step explanation:

A quadratic equation has one root if the discriminant is 0.

That is we need b^2-4ac=0 for this particular question.

Compare the following to find a,b, \text{ and } c:

ax^2+bx+c=0

my^2+2y-4=0

The variable x is representative of the variable y here.

a=m

b=2

c=-4

Plug in into b^2-4ac=0:

(2)^2-4(m)(-4)=0

4+16m=0

Subtract 4 on both sides:

16m=-4

Divide both sides by 16:

m=\frac{-4}{16}

Reduce:

m=\frac{-1}{4}

6 0
2 years ago
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Helpppppppppp there are 2 pics attached
fiasKO [112]

Answer:

Option A

Option B

Option E

Option G

Option A and E go together

and

Option B and G go together.

4 0
3 years ago
F(x) = –3x2 – 7x<br> Find f(7)
Dominik [7]

Answer:

ez

Step-by-step explanation:

7 0
2 years ago
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Which are the solutions of the quadratic equation? x2 = –5x – 3 –5, 0 StartFraction negative 5 minus StartRoot 13 EndRoot Over 2
boyakko [2]

<u>Given</u>:

The quadratic equation is x^{2}=-5 x-3

We need to determine the solutions of the quadratic equation.

<u>Solution</u>:

Let us solve the equation to determine the value of x.

Adding both sides of the equation by 5x and 3, we get;

x^{2}+5 x+3=0

The solution of the equation can be determined using quadratic formula.

Thus, we get;

x=\frac{-5 \pm \sqrt{5^{2}-4 \cdot 1 \cdot 3}}{2 \cdot 1}

x=\frac{-5 \pm \sqrt{25-12}}{2 }

x=\frac{-5 \pm \sqrt{13}}{2 }

Thus, the two roots of the equation are x=\frac{-5 + \sqrt{13}}{2 } and x=\frac{-5- \sqrt{13}}{2 }

Hence, the solutions of the equation are x=\frac{-5 + \sqrt{13}}{2 } and x=\frac{-5- \sqrt{13}}{2 }

6 0
3 years ago
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