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sasho [114]
3 years ago
15

Is anyone here familiar with Algebra 1B problems?

Mathematics
1 answer:
NISA [10]3 years ago
6 0

Answer:

Part 5) 3

Part 6) 2

Par 7) S=5\ ft

Part 8) S=7\ m

Step-by-step explanation:

Part 5) we have

\sqrt[3]{27}

we know that

27=3^{3}

substitute

\sqrt[3]{27}=\sqrt[3]{3^{3}}=3^{\frac{3}{3}}=3

Part 6) we have

\sqrt[4]{16}

we know that

16=2^{4}

substitute

\sqrt[4]{16}=\sqrt[4]{2^{4}}=2^{\frac{4}{4}}=2

Part 7) we know that

The volume of the cube is equal to

V=S^{3}

where

S is the length side of the cube

In this problem we have

V=125\ ft^{3}

substitute

125=S^{3}

S=\sqrt[3]{125}=\sqrt[3]{5^{3}}=5^{\frac{3}{3}}=5\ ft

Part 8) we know that

The volume of the cube is equal to

V=S^{3}

where

S is the length side of the cube

In this problem we have

V=343\ m^{3}

substitute

343=S^{3}

S=\sqrt[3]{343}=\sqrt[3]{7^{3}}=7^{\frac{3}{3}}=7\ m

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Answer:

12x + 2

Step-by-step explanation:

Let the number be represented by x.

Then five times the number = 5*x

Seven times the number = 7*x

Sum of 5 times the number minus -2 = \[5*x - (-2)\] = \[5x +2\]

Adding seven times the number to this expression yields, \[5x+2+7x\]

\[= (5+7)x+2\]

\[= 12x+2\]

So the simplified expression corresponds to 12x + 2.

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3 years ago
Write an equation for a circle with a diameter that has endpoints at (–10, 1) and (–8, 5). Round to the nearest tenth if necessa
satela [25.4K]

The standard form for the equation of a circle is :

<span><span><span> (x−h)^</span>2</span>+<span><span>(y−k)^</span>2</span>=<span>r2</span></span><span> ----------- EQ(1)
</span><span> where </span><span>handk</span><span> are the </span><span>x and y</span><span> coordinates of the center of the circle and </span>r<span> is the radius.
</span> The center of the circle is the midpoint of the diameter.

 So the midpoint of the diameter with endpoints at (−10,1)and(−8,5) is :

 ((−10+(−8))/2,(1+5)/2)=(−9,3)

 So the point (−9,3) is the center of the circle.

  Now, use the distance formula to find the radius of the circle:

  r^2=(−10−(−9))^2+(1−3)^2=1+4=5

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 Subtituting h=−9, k=3 and r=√5 into EQ(1) gives :

 (x+9)^2+(y−3)^2=5

7 0
3 years ago
Identify the range of the function shown in the graph.<br><br> help
amm1812

Answer:

-5 ≤ y ≤ 5

Step-by-step explanation:

maximum y value is 5

minimum y value is -5

-5 ≤ y ≤ 5

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tino4ka555 [31]

Answer:

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Step-by-step explanation:

4 0
3 years ago
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Aleksandr-060686 [28]

Answer:

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Step-by-step explanation:

correlation, r = \frac{∑(y-\alpha)(x-\beta  }{\sqrt{∑(y-\alpha)^2∑(x-\beta )^2 } }

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This represents a weak negative correlation

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