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2 years ago

Select the locations on the number line to plot the points 9/2 and −7/2 .

Mathematics

2 answers:

ludmilkaskok [199]2 years ago

8 0

Answer:

The easiest way to plot the fraction in the number line is to change the fraction into the decimal then plot it, as decimal is easily detected in the number line as compare to the fraction.

The Fraction $\frac{9}{2} = 4.5$

and value 4.5 lies exact between 4 and 5.

The Fraction $\frac{-7}{2} = -3.5$

and value 4.5 lies exact between -4 and -3.

It is shown below.

Alex_Xolod [135]2 years ago

3 0

9/2 would be on the right -7/2 will be on the left hope i helped

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Manuel buys 7.9 pounds of strawberries. Each pound costs $3.89. Which expression gives the best estimate of the cost of the stra

topjm [15]

The options you gave are extremely confusing, but if you round 7.9 it is 8, and if you round $3.89 it is $4. Therefore, the best estimate would be 8•$4, or $32

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choli [55]

Answer:

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Step-by-step explanation:

The scale is proportional to the actual:

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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.

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2 years ago

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tiny-mole [99]

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Lilit [14]

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