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2 years ago

15

What is the value of k in the product of powers below? 10 Superscript negative 3 Baseline times 10 times 10 Superscript k Baseli

ne = 10 Superscript negative 3 Baseline = StartFraction 1 Over 10 cubed EndFraction Negative 3 Negative 1 0 1

2 answers:

poizon [28]2 years ago

7 0

Answer: -1

Step-by-step explanation:

10^-3 x 10 = 10^-2

But you need it to be 1/10^3

So 10^-2 x 10^-1 = -2 + -1

= 10^-3

In order to put it in simplest form, the power needs to be positive.

So = 1/10^3

Meaning the missing power is -1

Karolina [17]2 years ago

6 0

Answer:

-1

Step-by-step explanation:

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On a vacation in Puerto Rico, Rose jumped off a cliff into a river in EL Yunque Forest Reserve. Her height as a function of time

balandron [24]

H (t) = - 16t ^ 2 + 16t + 480
For this case, the first thing to do is to match the polynomial to zero to find the roots.
We have then:
0 = -16t ^ 2 + 16t + 480
From here, we obtain the following roots of the polynomial:
t1 = -5
t2 = 6
We ignore the negative root because the time is always greater than zero.
Answer:
it takes Rose to hit the water about
t = 6 seconds

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2 years ago

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The mean gross annual incomes of certified welders are normally distributed with the mean of $20,000 and a standard deviation of

grandymaker [24]

Answer:

At the 0.10 level of significance the z table gives critical values of -1.645 and 1.645 for two-tailed test.

Step-by-step explanation:

We are given that the mean gross annual incomes of certified welders are normally distributed with the mean of $20,000 and a standard deviation of $2,000.

The ship building association wishes to find out whether their welders earn more or less than $20,000 annually.

<u><em>Let </em></u> $\mu$ <u><em> = mean gross annual incomes of certified welders</em></u>

So, Null Hypothesis, $H_0$ : $\mu$ = $20,000

Alternate hypothesis, $H_A$ : $\mu \neq$ $20,000

Here, null hypothesis states that the mean income of welders is equal to $20,000.

On the other hand, alternate hypothesis states that the mean income of welders is not $20,000.

Also, the test statistics that would be used here is <u>One-sample z test</u> <u>statistics</u> as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean income

$\sigma$ = population standard deviation = $2,000

n = sample size

Now, at the 0.10 level of significance the z table gives critical values of -1.645 and 1.645 for two-tailed test.

3 0

2 years ago

If point C is inside AvB, then __+m CvB= m AvB

adelina 88 [10]

Answer: OPTION A

Step-by-step explanation:

It is important to remember the "Angle Addition Postulate" in order to solve this exercise.

You have this larger angle:

$\angle AVB$

According to the "Angle Addition Postulate", if the point "C" is in the interior of $\angle AVB$ , then the measure of this angle can be obtained by adding the angle $\angle AVC$ and the angle $\angle CVB$ .

You can observe in the figure that:

$m\angle AVC=39\°\\\\\mangle CVB=23\°$

Then you get that:

$m\angle AVC+m\angle CVB=m\angle AVB\\\\39\°+23\°=62\°$

Therefore, the correct answer is Option A.

3 0

2 years ago

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Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0

2 years ago

Myra won the hurdles event for the seventh graders. She completed the race in 22.8 seconds and knocked down 1 hurdle. Byron won

Nadya [2.5K]

Answer:

Myra: 24.8s (winner)

Byron: 26.3s

Step-by-step explanation:

Myra knocked down 1, so therefore there are 2 points added onto her time and it is now 24.8s.

Byron knocked down 3, so there are 6 points added on for 2 seconds per each hurdle. His time is now 26.3

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2 years ago