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2 years ago

You have 2 cats and 3 dogs. Write the ratio of cats to dogs two different ways.

Mathematics

1 answer:

fomenos2 years ago

3 0

Answer:

2:3

3:2

Step-by-step explanation:

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Mark, June & Henry share some money in the ratio 1 : 9 : 4.

Sladkaya [172]

45

Mark brainliest please

Hope this helps

4 0

2 years ago

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Can someone please help I’m very confused

Oxana [17]

y + 6 = 4/3(x - 2)

First, distribute the 4/3

y + 6 = 4/3x - 8/3

Subtract 6 from both sides.

y = 4/3x - 26/3

-26/3 is the y-intercept so start the graph at coordinate (0, -26/3)

After you plot the y-intercept, add 4 to they y-coordinate and 3 to the x-coordinate of the y-intercept to get the next point.

0 + 3 = 3

-26/3 + 4 = -14/3

The next point should be at (3, -14/3)

Add as many points as your professor requires.

8 0

2 years ago

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Which expression is equivalent to five to the tenth power multiplied by five to the fifth

ioda

Answer:

30517578125

Step-by-step explanation:

do u mean multiplying it? if not then plz lmk so i could help

6 0

3 years ago

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Find derivative problem<br> Find B’(6)

dalvyx [7]

Answer:

$B^\prime(6) \approx -28.17$

Step-by-step explanation:

We have:

$\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)$

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

$\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]$

We can move the constant outside:

$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

$(uv)^\prime=u^\prime v+u v^\prime$

We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$

Therefore:

$\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]$

By simplification:

$\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17$

So, the slope of the tangent line to the point (6, B(6)) is -28.17.

5 0

2 years ago

Which value of x makes the equation true? x – 5.1 = –7.6

ASHA 777 [7]

X-5.1 = -7.6
+5.1 +5.1
X= -2.5

7 0

3 years ago

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