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3 years ago

PLZZZZZZZZZZZZZZZZ HELP

Mathematics

1 answer:

Slav-nsk [51]3 years ago

5 0

Answer:

The first box in the girl row is 18, the second one in the boy row is 8, and the bottom one is also 18.

Step-by-step explanation:

To get 81 to 9, you have to divide by 9, which means you have to do the same to 72. You get the ration 9 to 8, 9 being girls and 8 being guys. Then to get from 8 guys to 16, you have to multiply by 2, which means you do the same to 9 and get the answer of 18 girls. Hope this helps :)

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Prove that :( 1 + 1/<img src="https://tex.z-dn.net/?f=tan%5E%7B2%7DA" id="TexFormula1" title="tan^{2}A" alt="tan^{2}A" align="ab

Aliun [14]

Answer:

See explanation

Step-by-step explanation:

Simplify left and right parts separately.

$\left(1+\dfrac{1}{\tan^2A}\right)\left(1+\dfrac{1}{\cot ^2A}\right)\\ \\=\left(1+\dfrac{1}{\frac{\sin^2A}{\cos^2A}}\right)\left(1+\dfrac{1}{\frac{\cos^2A}{\sin^2A}}\right)\\ \\=\left(1+\dfrac{\cos^2A}{\sin^2A}\right)\left(1+\dfrac{\sin^2A}{\cos^2A}\right)\\ \\=\dfrac{\sin^2A+\cos^2A}{\sin^2A}\cdot \dfrac{\cos^2A+\sin^A}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A}\cdot \dfrac{1}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$

<u>Right part:</u>

$\dfrac{1}{\sin^2A-\sin^4A}\\ \\=\dfrac{1}{\sin^2A(1-\sin^2A)}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$

Since simplified left and right parts are the same, then the equality is true.

3 0

3 years ago

Simplify the following

den301095 [7]

Answer:

1) 11 $\sqrt{3}$

2) 2 $\sqrt{2}$

3) $20\sqrt{3} + 15\sqrt{2}$

4) $53 + 12\sqrt{10}$

5) -2

6) $7\sqrt{2} - 5\sqrt{3}$

Step-by-step explanation:

1) 2 $\sqrt{12}$ + 3 $\sqrt{48}$ - $\sqrt{75}$

=(2 × 2 $\sqrt{3}$ )+ (3 × 4 $\sqrt{3}$ ) - 5 $\sqrt{3}$

= 4 $\sqrt{3}$ + 12 $\sqrt{3}$ - 5 $\sqrt{3}$

= 11 $\sqrt{3}$

2) 4 $\sqrt{8}$ -2 $\sqrt{98}$ + $\sqrt{128}$

= (4 × 2 $\sqrt{2}$ ) - (2 × 7 $\sqrt{2}$ ) + 8 $\sqrt{2}$

= 8 $\sqrt{2}$ - 14 $\sqrt{2}$ +8 $\sqrt{2}$

= 2 $\sqrt{2}$

3) 5 $\sqrt{12\\}$ - 3 $\sqrt{18} + 4 \sqrt{72} +2\sqrt{75}$

= 5× $2\sqrt{3}$ - 3× $3\sqrt{2}$ + 4× $6\sqrt{2}$ + 2× $5\sqrt{3}$

= $10\sqrt{3} - 9\sqrt{2} +24\sqrt{2} +10\sqrt{3}$

= $20\sqrt{3} + 15\sqrt{2}$

4) $(2\sqrt{2} + 3\sqrt{5} )^{2}$

= $8 + 12\sqrt{10} + 45$

= $53 + 12\sqrt{10}$

5) $(1+\sqrt{3} ) (1-\sqrt{3} )$

= $1 - 3$

= -2

6) $(2\sqrt{6} -1) (\sqrt{3} -\sqrt{2} )$

= $2\sqrt{18}-2\sqrt{12} -\sqrt{3} +\sqrt{2}$

= 2× $3\sqrt{2}$ - 2× $2\sqrt{3}$ - $\sqrt{3} + \sqrt{2}$

= $6\sqrt{2} - 4\sqrt{3} -\sqrt{3} +\sqrt{2}$

= $7\sqrt{2} - 5\sqrt{3}$

Hope the working out is clear and will help you. :)

5 0

2 years ago

Read 2 more answers

Divide the polynomial (x^2 + 3x + 5) by 2x + 1

umka21 [38]

Answer:

50888888888888888888+++++

7 0

3 years ago

Find the x-intercept of the parabola with vertex (1,-13) and y-intercept (0,-11).

Advocard [28]

1. find equation

general solutions are:
$y = a x^{2} +bx +c \\ \\ \frac{dy}{dx} = 2ax +b$

for P(0,-11)

$-11 = 0a + 0b +c$ ⇒ $c = -11$

for p(1,-13)
$-13 = 1a + 1b -11 \\ a+b = -2$

and

$0 = 2a + b \\ b = -2a$

solve for a and b:
$a - 2a = -2 \\ a = 2 \\ b = -4$

the total equation is now:

$y = 2 x^{2} -4x -11$

To find the x-intercept set y=0 and solve for x

$0 = 2 x^{2} -4x-11$

4 0

3 years ago

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PLEASE HELP IF YOU ARE GOOD AT ALGEBRA 2 MATH

REY [17]

Answer:I think a is:78.3 since it would be the same and since 0 doesn't have a value.

I think b is: 139.5 because all you are doing is subtracting 287.1-147.6

7 0

3 years ago