Write the answer in interval notation

Sketch the region enclosed by x^2+y^2=6 and x+y=0. decide whether to integrate with respect to x or y. then find the area of the

kondor19780726 [428]

It's 3 beacause you add all of them together to get 0

4 0

3 years ago

What is 6.34 when the 34 is repeating as a fraction

lana [24]

ANSWER:

6_34/99

STEP:

So yes. When a decimal is repeating, you can take the repeating number (most likely a decimal) and put 99 under it. Since 99 cannot be solved, you put 99. So, 34/99. Though we are not finished. There is still the whole 6 number left. So, you do 6_34/99.

Proof:

10x=6.6...

-x=-0.6...

9x=6

x=6/9=1/3.

3 0

3 years ago

Analysis of an accident scene indicates a car was traveling at a velocity of 69.5 mph (31.1 m/s) along the positive x-axis at th

STatiana [176]

We can find the acceleration via

${v_f}^2-{v_i}^2=2a\Delta x$

We have

$\left(5.20\dfrac{\rm m}{\rm s}\right)^2-\left(31.1\dfrac{\rm m}{\rm s}\right)^2=2a(115\,\mathrm m)$

$\implies\boxed{a=-4.09\dfrac{\rm m}{\mathrm s^2}}$

Then by definition of average acceleration,

$a_{\rm ave}=\dfrac{v_f-v_i}t$

so that

$-4.09\dfrac{\rm m}{\mathrm s^2}=\dfrac{5.20\frac{\rm m}{\rm s}-31.1\frac{\rm m}{\rm s}}t$

$\implies\boxed{t=6.33\,\mathrm s}$

We alternatively could have found the time without knowing the acceleration. Since acceleration is constant, the average velocity is

$v_{\rm ave}=\dfrac{x_f-x_i}t=\dfrac{v_f+v_i}2$

Then

$\dfrac{115\,\rm m}t=\dfrac{5.20\frac{\rm m}{\rm s}+31.1\frac{\rm m}{\rm s}}2$

$\implies\boxed{t=6.33\,\mathrm s}$

7 0

3 years ago

W = p ( y -z ) Please solve this for z.

11111nata11111 [884]

w = p(y - z)

divide p on both sides

$\frac{w}{p} = y - z$

Subtract y on both sides

$\frac{w}{p} - y = -z$

Divide -1 on both sides

$-\frac{w}{p} +y=z$

8 0

3 years ago

A worker was paid a salary of $10,500 in 1985. Each year, a salary increase of 6% of the previous year's salary was awarded. How

Mazyrski [523]

Note that 6% converted to a decimal number is 6/100=0.06. Also note that 6% of a certain quantity x is 0.06x.

Here is how much the worker earned each year:

In the year 1985 the worker earned <span>$10,500.

</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:

$10,500(1+0.06).

In the year 1987 the worker earned

$10,500(1+0.06) + 0.06[$10,500(1+0.06)].

Now we can factorize $10,500(1+0.06) and write the earnings as:

$10,500(1+0.06) [1+0.06]= $$10,500(1.06)^2$ .

Similarly we can check that in the year 1987 the worker earned $$10,500(1.06)^3$ , which makes the pattern clear.

We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:

$10,500+10,500(1.06)^1+10,500(1.06)^2+10,500(1.06)^3...10,500(1.06)^{26}$ .

Factorizing, we have

$=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]$

We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula

$\sum_{i=1}^{n} a_i= a(\frac{1-r^n}{1-r})$ (where a is the first term and r is the common ratio) we have:

$\sum_{i=1}^{26} a_i= 1(\frac{1-(1.06)^{26}}{1-1.06})= \frac{1-4.55}{-0.06}= 59.17$ .

Finally, multiplying 10,500 by 59.17 we have 621.285 ($).

The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating $(1.06)^{26}$ .

Answer: D

4 0

3 years ago