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3 years ago

The ratio of boys and girls at the beach cleanup was 7:8. If there were 42 boys, how much girls were there

Mathematics

1 answer:

ratelena [41]3 years ago

3 0

Answer:

48 girls

Step-by-step explanation:

Divide 42 by 7 and then multiply the answer by 8. So 42/7 is 6 and 6 multiplied by 8 is 48 so if there are 42 boys there are 48 girls.

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In a sample of 100 students, 30 wear size medium t-shirts. Use a proportion to calculate approximately how many medium t-shirts

Talja [164]

Answer:

282 students

Step-by-step explanation:

7 0

3 years ago

Tan^2 x+sec^2 x=1 for all values of x.<br> True or False

gogolik [260]

False

because tan^2 45 = 1^2 = 1
and sec^2 x = 1/ cos^2 45 = (sqrt2)^2 = 2

so the sum = 3

3 0

2 years ago

Read 2 more answers

John is creating a Thanksgiving display at the store where he works, using only canned pumpkin and canned green beans. He needs

motikmotik

Answer:

Step-by-step explanation:

The ratio of green beans (g) to pumpkin (p) is ...

... g : p = 1 : 5

Then the ratio of green beans to the total is ...

... g : (g+p) = 1 : (1+5) = 1 : 6

Since you have

... g/total = (1/6)

... g = total · 1/6

you can substitute 114 for the total to find ...

... g = 114 · 1/6 = 19

6 0

3 years ago

A sphere has a volume of V=900 in cubed. What is the surface area?

ycow [4]

First, we must solve for the radius of the sphere:
V= $\frac{4}{3}$ $\pi$ $r^{3}$
r=(3 $\frac{V}{4 \pi }$ ) $^{ \frac{1}{3} }$
r=(3* $\frac{900}{4 \pi }$ ) $^{ \frac{1}{3} }$
r≈5.99

Second, we must solve for surface area:
A=4 $\pi$ $r^{2}$
A=4* $\pi$ * $5.99^{2}$
A≈450.88 $in^{2}$

4 0

3 years ago

What is the derivative of x times squaareo rot of x+ 6?

Dafna1 [17]

Hey there, hope I can help!

$\mathrm{Apply\:the\:Product\:Rule}: \left(f\cdot g\right)^'=f^'\cdot g+f\cdot g^'$
$f=x,\:g=\sqrt{x+6} \ \textgreater \ \frac{d}{dx}\left(x\right)\sqrt{x+6}+\frac{d}{dx}\left(\sqrt{x+6}\right)x \ \textgreater \ \frac{d}{dx}\left(x\right) \ \textgreater \ 1$

$\frac{d}{dx}\left(\sqrt{x+6}\right) \ \textgreater \ \mathrm{Apply\:the\:chain\:rule}: \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx} \ \textgreater \ =\sqrt{u},\:\:u=x+6$
$\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(x+6\right)$

$\frac{d}{du}\left(\sqrt{u}\right) \ \textgreater \ \mathrm{Apply\:radical\:rule}: \sqrt{a}=a^{\frac{1}{2}} \ \textgreater \ \frac{d}{du}\left(u^{\frac{1}{2}}\right)$
$\mathrm{Apply\:the\:Power\:Rule}: \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1} \ \textgreater \ \frac{1}{2}u^{\frac{1}{2}-1} \ \textgreater \ Simplify \ \textgreater \ \frac{1}{2\sqrt{u}}$

$\frac{d}{dx}\left(x+6\right) \ \textgreater \ \mathrm{Apply\:the\:Sum/Difference\:Rule}: \left(f\pm g\right)^'=f^'\pm g^'$
$\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(6\right)$

$\frac{d}{dx}\left(x\right) \ \textgreater \ 1$
$\frac{d}{dx}\left(6\right) \ \textgreater \ 0$

$\frac{1}{2\sqrt{u}}\cdot \:1 \ \textgreater \ \mathrm{Substitute\:back}\:u=x+6 \ \textgreater \ \frac{1}{2\sqrt{x+6}}\cdot \:1 \ \textgreater \ Simplify \ \textgreater \ \frac{1}{2\sqrt{x+6}}$

$1\cdot \sqrt{x+6}+\frac{1}{2\sqrt{x+6}}x \ \textgreater \ Simplify$

$1\cdot \sqrt{x+6} \ \textgreater \ \sqrt{x+6}$
$\frac{1}{2\sqrt{x+6}}x \ \textgreater \ \frac{x}{2\sqrt{x+6}}$
$\sqrt{x+6}+\frac{x}{2\sqrt{x+6}}$

$\mathrm{Convert\:element\:to\:fraction}: \sqrt{x+6}=\frac{\sqrt{x+6}}{1} \ \textgreater \ \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}}{1}$

Find the LCD
$2\sqrt{x+6} \ \textgreater \ \mathrm{Adjust\:Fractions\:based\:on\:the\:LCD} \ \textgreater \ \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}\cdot \:2\sqrt{x+6}}{2\sqrt{x+6}}$

$Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions$
$\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{x+2\sqrt{x+6}\sqrt{x+6}}{2\sqrt{x+6}}$

$x+2\sqrt{x+6}\sqrt{x+6} \ \textgreater \ \mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c}$
$\sqrt{x+6}\sqrt{x+6}=\:\left(x+6\right)^{\frac{1}{2}+\frac{1}{2}}=\:\left(x+6\right)^1=\:x+6 \ \textgreater \ x+2\left(x+6\right)$
$\frac{x+2\left(x+6\right)}{2\sqrt{x+6}}$

$x+2\left(x+6\right) \ \textgreater \ 2\left(x+6\right) \ \textgreater \ 2\cdot \:x+2\cdot \:6 \ \textgreater \ 2x+12 \ \textgreater \ x+2x+12$
$3x+12$

Therefore the derivative of the given equation is
$\frac{3x+12}{2\sqrt{x+6}}$

Hope this helps!

8 0

2 years ago