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2 years ago

There are 81 red, blue and yellow

2 answers:

5 0

You take the 81 counters and pull 9 to the side and that’s 72. take that 72 you divide that by 2 and get 36. so it’s 36 separate blue and yellow counters. there is somehow 9 more red than yellow. which I do not get lol but I did that good.

wel2 years ago

3 0

Step-by-step explanation:

Here Given

Total counters = 81

red counters are 9 more than the yellow ones.

And Yellow counter = blue counter

So let the Yellow counter be x and as blue counter and yellow counter are same then blue counter will be also x

And by the question

Red counter = 9 + x

Now

9 + x + x + x = 81

9 + 3x = 81

3x = 81 - 9

3x = 72

x = 72 / 3

Therefore x = 24

Now

Red counters = 9 + 24 = 33

Blue counters = 24

Yellow counters = 24

Hope it helped :)

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Determine the slope of the line that contains the given points <br><br> J(-5, -2), K(5, −4)

Irina-Kira [14]

Answer:

$-\frac15$

Step-by-step explanation:

Hello!

We can utilize the slope formula to find the slope.

Slope Formula: $S = \frac{y_2-y_1}{x_2-x_1}$

Remember that a coordinate is written in the form (x,y)

<h3>Find the Slope</h3>

$S = \frac{y_2-y_1}{x_2-x_1}$
$S = \frac{-4-(-2)}{5-(-5)}$
$S = \frac{-2}{10}$
$S = -\frac15$

The slope of the line is $-\frac15$ .

4 0

1 year ago

Read 2 more answers

Evaluate Dx / ^ 9-8x - x2^

Solnce55 [7]

It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for $\sqrt x$ .

In that case, you want to find the antiderivative,

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}$

Complete the square in the denominator:

$9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2$

Now substitute $x+4=5\sin y$ , so that $\mathrm dx=5\cos y\,\mathrm dy$ . Then

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy$

which simplifies to

$\displaystyle\int\frac{5\cos 
y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy$

Now, recall that $\sqrt{x^2}=|x|$ . But we want the substitution we made to be reversible, so that

$x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)$

which implies that $-\dfrac\pi2\le y\le\dfrac\pi2$ . (This is the range of the inverse sine function.)

Under these conditions, we have $\cos y\ge0$ , which lets us reduce $\sqrt{\cos^2y}=|\cos y|=\cos y$ . Finally,

$\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C$

and back-substituting to get this in terms of $x$ yields

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C$

4 0

2 years ago

Jeremy is taking a photography class, but he doesn't own a camera. The class organizers will rent him a camera for $6 per day. J

Mazyrski [523]

Solving the given inequality for d, we get...

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6d < 35
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d < 5.83

Which means that d can be any of the values in this set: {0, 1, 2, 3, 4, 5}

The smallest d can be is 0. In this scenario, Jeremy pays the $15 registration but doesn't rent the camera at all
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Notice how I'm rounding down regardless how close 5.83 is to 6

8 0

2 years ago

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KATRIN_1 [288]

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2 years ago

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shusha [124]

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