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3 years ago

In 36 days a girl spent $18 on average what did she spend Daly

Mathematics

1 answer:

Yuri [45]3 years ago

8 0

5 cent, how I got my answer.

I simply divided $18÷36=0.5

I hope this helps!

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The equation shows a number multiplied by 5. n x 5=? Which of these is true about the results?

stealth61 [152]

Answer:

Step-by-step explanation:

Michelle collected data on the weight of 200 apples at the farmer's market. Most weighed 0.30 –.40 pounds, but three weighed over a pound each. Michelle identified this as a striking deviation. What should she do with it?

urrence.

8 0

3 years ago

Which sentence is correct?A.B.C.D.

Dennis_Churaev [7]

Answer: B is correct

Step-by-step explanation:

if you do 2 ÷ 5 you get 0.4. But that isn't an option. So, the only other possibility would be -2 ÷ 5 which gives us -0.4, and that makes sense.

I'm not the greatest at explaining this one but hopefully, this helps!

3 0

3 years ago

The equation for the line of best fit for a data set is y = 2x + 1.5. If the point (1, 4) is in the data set, what is the residu

Andreas93 [3]

A. 0.5 well at least that's the answer in ingenuity.<span />

5 0

3 years ago

Read 2 more answers

Find the missing length. Give all answers in the simplest radical form.

Solnce55 [7]

Answer:

x = 2 $\sqrt{74}$

Step-by-step explanation:

Since the triangle is right use Pythagoras' identity to solve for x

The square on the hypotenuse is equal to the sum of the squares on the other two sides, that is

x² = 14² + 10² = 196 + 100 = 296

Take the square root of both sides

x = $\sqrt{296}$ = 2 $\sqrt{74}$

8 0

3 years ago

Find the complex fourth roots \[-\sqrt{3}+\iota \] in polar form.

____ [38]

Let $z=-\sqrt3+i$ . Then

$|z|=\sqrt{(-\sqrt3)^2+1^2}=2$

$z$ lies in the second quadrant, so

$\arg z=\pi+\tan^{-1}\left(-\dfrac1{\sqrt3}\right)=\dfrac{5\pi}6$

So we have

$z=2e^{i5\pi/6}$

and the fourth roots of $z$ are

$2^{1/4}e^{i(5\pi/6+k\pi)/4}$

where $k\in\{0,1,2,3\}$ . In particular, they are

$2^{1/4}e^{i(5\pi/6)/4}=2^{1/4}e^{i5\pi/24}$

$2^{1/4}e^{i(5\pi/6+2\pi)/4}=2^{1/4}e^{i17\pi/24}$

$2^{1/4}e^{i(5\pi/6+4\pi)/4}=2^{1/4}e^{i29\pi/24}$

$2^{1/4}e^{i(5\pi/6+6\pi)/4}=2^{1/4}e^{i41\pi/24}$

7 0

3 years ago