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3 years ago

X + 2y = 7 x + 5 = 3y What is the resulting equation?

Mathematics

1 answer:

levacccp [35]3 years ago

5 0

Download on your phone photo math it scans the problem and solves it for u

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A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat,

Paraphin [41]

Answer:

3/7.

Step-by-step explanation:

The total number of ways to pick 4 out of 8 = 8C4

= 8!/ 4!4!

= 70 ways.

The number of ways when 2 are women = 5C2 * 3

= 10 * 3

= 30.

So the required probability = 30/70

= 3/7.

6 0

3 years ago

Latifah drew this obtuse, scalene triangle.

kumpel [21]

Quadrilateral that is a parallelogram

7 0

3 years ago

Sheila purchased ski equipment for $513 using a six-month deferred payment plan. The interest rate after the introductory period

klemol [59]

$65 * 6 = $390 total monthly payments made

The downpayment made is $125, so the total balance at the end is:

balance = $513 – ($125 + $390)

balance = - $2

<span>There is an excess of 2 dollars at the end.</span>

4 0

3 years ago

In a certain year the population of a country reached 321 million. The overall birth rate was births per 1000, and the overall

kkurt [141]

Answer:

Then it would stay at 321 million

Step-by-step explanation:

if the birth and death rate is both 1000 then nothing will change. Population wise.

4 0

3 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.

5 0

3 years ago