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2 years ago

What is 3a + 3/6. if a = 1/4

Mathematics

1 answer:

Sav [38]2 years ago

7 0

Answer:

5/4

Step-by-step explanation:

Solve by plugging in the value to "a".

Solve:

3a + 3/6

3(1/4) + 3/6

Multiply the parenthesis.

3/4 + 3/6

Simplify 3/6

3/4 + 1/2

Find the common denominator, which is 4. Make the denominators the same.

3/4 + 2/4

Add the fractions.

5/4

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Step-by-step explanation:

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3 years ago

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2 years ago

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.

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3 years ago

PLS HELP ASAP FIRST CORRECT ANSWER GETS BRAINLEIST

solmaris [256]

Answer:

50 cm^2

Step-by-step explanation:

area of rectangle = base * height

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2 years ago

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kicyunya [14]

Answer:

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Step-by-step explanation:

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7 0

3 years ago

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