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3 years ago

-5.25 x -6.4 Pls help!

Mathematics

2 answers:

xz_007 [3.2K]3 years ago

8 0

Answer:

33.6

Step-by-step explanation:

weeeeeb [17]3 years ago

5 0

Answer:

33.6

Step-by-step explanation:

please pick me as brainliest thanks!

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Step-by-step explanation:

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3 years ago

Read 2 more answers

You purchased a rare painting for $150 that is increasing in value by 3% annually. How many years will it take until it is doubl

Ahat [919]

Answer:

23 years.

Step-by-step explanation:

It is given that the initial price of painting is $150 and its values increasing by 3% annually.

We need to find how many years will it take until it is doubled in value.

The value of painting after t years is given by

$y=150(1+0.03)^t$

$y=150(1.03)^t$

The value of painting after double is 300. Substitute y=300.

$300=150(1.03)^t$

Divide both sides by 150.

$2=(1.03)^t$

Taking log both sides.

$\log 2=\log (1.03)^t$

$\log 2=t\log (1.03)$

$t=\dfrac{\log 2}{\log (1.03)}$

$t=23.44977$

$t\approx 23$

Therefore, the required number of years is 23.

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3 years ago

9x4y6 + 42x5y5 +49x3y2

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3 years ago

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3 years ago

PLEASE HELP ME IVE POSTED THIS 765678 AND STILL NO RESPONSE

Debora [2.8K]

Problem 1

<h3>Answer: 6.7</h3>

----------------

Work Shown:

The two points are $(x_1,y_1) = (1,-2)$ and $(x_2,y_2) = (4,4)$

Apply the distance formula to get the following

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-4)^2 + (-2-4)^2}\\\\d = \sqrt{(-3)^2 + (-6)^2}\\\\d = \sqrt{9 + 36}\\\\d = \sqrt{45}\\\\d \approx 6.7082039\\\\d \approx 6.7\\\\$

The distance between the two endpoints is roughly 6.7 units. This is the same as saying the segment is roughly 6.7 units long.

======================================================

Problem 2

<h3>Answer: 3.6</h3>

----------------

Work Shown:

We'll use the distance formula here as well.

This time we have the two points $(x_1,y_1) = (3,1)$ and $(x_2,y_2) = (5,-2)$

The distance between them is...

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-5)^2 + (1-(-2))^2}\\\\d = \sqrt{(3-5)^2 + (1+2)^2}\\\\d = \sqrt{(-2)^2 + (3)^2}\\\\d = \sqrt{4 + 9}\\\\d = \sqrt{13}\\\\d \approx 3.6055513\\\\d \approx 3.6\\\\$

This distance is approximate.

5 0

3 years ago