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shutvik [7]
4 years ago
11

WILL GIVE BRAINLIEST

Mathematics
2 answers:
Degger [83]4 years ago
8 0
Based on the photo The answers would be D The price of stock over time
olganol [36]4 years ago
4 0
I think it D




hope it helps
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Angle $eab$ is a right angle, and $be = 9$ units. what is the number of square units in the sum of the areas of the two squares
Alla [95]
Let the measure of side AB be x, then, the measue of side AE is given by

AE=\sqrt{9^2-x^2}.

Now, ABCD is a square of size x, thus the area of square ABCD is given by

Area=x^2

Also, AEFG is a square of size \sqrt{9^2-x^2}, thus, the area of square AEFG is given by

Area=\left(\sqrt{9^2-x^2}\right)^2=9^2-x^2=81-x^2

<span>The sum of the areas of the two squares ABCD and AEFG is given by

x^2+81-x^2=81

Therefore, </span>the number of square units in the sum of the areas of the two squares <span>ABCD and AEFG is 81 square units.</span>
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3 years ago
Help pls i will give you brainlist but make sure its correct help ASAP
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It is 1,6 dollars for each pendant :)
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3 years ago
Liquid A has a density of 0.7g/cm^3
TiliK225 [7]
\rho = \frac{m}{V} \Rightarrow V=\frac{m}{\rho}
ρ - density, m - mass, V - volume

Liquid A:
\rho=0.7 \ \frac{g}{cm^3} \\&#10;m=140 \ g \\&#10;\\ V=\frac{140 \ g}{0.7 \ \frac{g}{cm^3}}=200 \ cm^3

Liquid B:
\rho=1.6 \ \frac{g}{cm^3} \\&#10;m=128 \ g \\&#10;\\ V=\frac{128 \ g}{1.6 \ \frac{g}{cm^3}}=80 \ cm^3

Liquid C:
m=140 \ g + 128 \ g=268 \ g \\&#10;V=200 \ cm^3 + 80 \ cm^3=280 \ cm^3 \\ \\&#10;\rho=\frac{268 \ g}{280 \ cm^3} \approx 0.96 \ \frac{g}{cm^3}

The density of liquid C is approximately 0.96 g/cm³.
5 0
3 years ago
Read 2 more answers
There are moles in 1.26 × 1024 particles.
EastWind [94]

Answer:

1.26 x 10^(24)

Step-by-step explanation:

hope this helped

6 0
3 years ago
On the teacher’s desk there were 5 boxes of pencils, j pencils in each. How many pencils were there on the desk?
sesenic [268]

Answer:

5j

Step-by-step explanation:

Multiply the number of boxes time the pencils in each box

5j

4 0
3 years ago
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