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3 years ago

10

the ratio of adults to students on the field trip is 3 to 5. if there are 15 students on the field trip, how many adults are the

re. please show how you got the answer please and thank you

2 answers:

nlexa [21]3 years ago

8 0

9 adults is the answer

Explanation: To go from 5 kids to 15 kids you have to multiply by 3, what you do to one side of the ratio you have to do to the other, so you take the 3 adults and also multiply it by 3, which gives you 9!

Aloiza [94]3 years ago

7 0

Answer:

There would be 9 adults.

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What is the explicit formula for the sequence below?<br><br> 4, -12, 36, -108 ...

Temka [501]

Answer:

This is a geometry sequence (I.e each term is found by multiplying the same value to the previous term).

T1 = 4

T2 = -12 = T1 X (-3) = 4 X (-3)

T3 = 36 = T2 X (-3) = 4 X (-3)^2

T4 = 108 = T3 X (-3) = 4 X (-3)^3

The general (explicit) formula for the n-th term of this sequence is:

S(n) = 4 X (-3)^(n-1)

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3 years ago

Which ordered pair could not be the coordinates for a clockwise rotation of the point W(-3, 4)?

Eduardwww [97]

Step-by-step explanation:

<em>Look at the picture.</em>

Any pair of choice can be the coordinates of the rotation of the point

W (-3, 4) clockwise.

All points have the same distance from the beginning.

The formula of a distance between the origin and a point (x, y):

$d=\sqrt{x^2+y^2}$

W(-3, 4)

$d=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

(3, -4)

$d=\sqrt{3^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5$

(4, 3)

$d=\sqrt{4^2+3^2}=\sqrt{9+16}=\sqrt{25}=5$

(-4, -3)

$d=\sqrt{(-4)^2+(-3)^2}=\sqrt{9+16}=\sqrt{25}=5$

(-4, 3)

$d=\sqrt{(-4)^2+3^2}=\sqrt{9+16}=\sqrt{25}=5$

3 0

3 years ago

Ples help me find slant assemtotes

FrozenT [24]

A polynomial asymptote is a function $p(x)$ such that

$\displaystyle\lim_{x\to\pm\infty}(f(x)-p(x))=0$

$(y+1)^2=4xy\implies y(x)=2x-1\pm2\sqrt{x^2-x}$

Since this equation defines a hyperbola, we expect the asymptotes to be lines of the form $p(x)=ax+b$ .

Ignore the negative root (we don't need it). If $y=2x-1+2\sqrt{x^2-x}$ , then we want to find constants $a,b$ such that

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0$

We have

$\sqrt{x^2-x}=\sqrt{x^2}\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=|x|\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=x\sqrt{1-\dfrac1x}$

since $x\to\infty$ forces us to have $x>0$ . And as $x\to\infty$ , the $\dfrac1x$ term is "negligible", so really $\sqrt{x^2-x}\approx x$ . We can then treat the limand like

$2x-1+2x-ax-b=(4-a)x-(b+1)$

which tells us that we would choose $a=4$ . You might be tempted to think $b=-1$ , but that won't be right, and that has to do with how we wrote off the "negligible" term. To find the actual value of $b$ , we have to solve for it in the following limit.

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-4x-b)=0$

$\displaystyle\lim_{x\to\infty}(\sqrt{x^2-x}-x)=\frac{b+1}2$

We write

$(\sqrt{x^2-x}-x)\cdot\dfrac{\sqrt{x^2-x}+x}{\sqrt{x^2-x}+x}=\dfrac{(x^2-x)-x^2}{\sqrt{x^2-x}+x}=-\dfrac x{x\sqrt{1-\frac1x}+x}=-\dfrac1{\sqrt{1-\frac1x}+1}$

Now as $x\to\infty$ , we see this expression approaching $-\dfrac12$ , so that

$-\dfrac12=\dfrac{b+1}2\implies b=-2$

So one asymptote of the hyperbola is the line $y=4x-2$ .

The other asymptote is obtained similarly by examining the limit as $x\to-\infty$ .

$\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0$

$\displaystyle\lim_{x\to-\infty}(2x-2x\sqrt{1-\frac1x}-ax-(b+1))=0$

Reduce the "negligible" term to get

$\displaystyle\lim_{x\to-\infty}(-ax-(b+1))=0$

Now we take $a=0$ , and again we're careful to not pick $b=-1$ .

$\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-b)=0$

$\displaystyle\lim_{x\to-\infty}(x+\sqrt{x^2-x})=\frac{b+1}2$

$(x+\sqrt{x^2-x})\cdot\dfrac{x-\sqrt{x^2-x}}{x-\sqrt{x^2-x}}=\dfrac{x^2-(x^2-x)}{x-\sqrt{x^2-x}}=\dfrac
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This time the limit is $\dfrac12$ , so

$\dfrac12=\dfrac{b+1}2\implies b=0$

which means the other asymptote is the line $y=0$ .

4 0

3 years ago

BRAINLIEST +STARSS!!!

yulyashka [42]

Answer:

The green would take 4 and the blue would take 16 times respectivly. I did it by finding the volumes of the cups in terms of pi, then dividing it by the volume of the sink.

Step-by-step explanation:

So the volume of the whole half sphere is 512pi.

Now we have to find the volumes of the cup.

Equation: $\pi r^2h$

Blue cup:

$\pi 2^2(8)\\\pi 4(8)\\\pi 32$

It would take 16 times to drain it completly.

Green cup:

$\pi 4^2(8)\\\pi 16(8)\\\pi 128$

It would take 4 times to drain it completly.

6 0

2 years ago

3(3x+9)=-5-2x what is x

-BARSIC- [3]

Answer:

x=2.9

Step-by-step explanation:

3(3x+9)= -5-2x ( distribute)

9x+27= -5-2x (add 2x) (subtract 27)

11x= 32 ( divide by 11)

x= 2.9

3 0

3 years ago