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3 years ago

13

The intersection of plane A and plane S will be

1 answer:

Darya [45]3 years ago

3 0

Answer: the intersection of plane A and plane S will be: line CW

the intersection of lines N and K is: point V

point X is the intersection of: line M and plane A

Step-by-step explanation:

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The length of a flower garden is 9 meters. The width of the garden, w, is unknown. If the area of the garden is greater

muminat

Answer:

The width should be more than 5 m. So it can be 6,7 ...

Step-by-step explanation:

The length of a flower garden is = 9 m

The width of a flower garden is = w

The area of a flower garden > 45 sq. m

The area of a flower garden = l * b

l * b > 45

9 * b > 45

the width should be more than 5 m. So it can be 6,7 ...

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3 years ago

What is -2m^2-2mn+8-10m+10n+mn+4n^2-5n simplify to

Mrac [35]

Find like terms and group them
3n and 5n are same and 4n^2 and 4n^2 are same
but 3mn and 5n are not and 3n and 2n^3 are not

group like terms
(-2m^2)+(-2mn+1mn)+(8)+(-10m)+(10n-5n)+(2n^2)
add like terms
(-2m^2)+(-mn)+(8)+(-10m)+(5n)+(2n^2)
-2m^2+2n^2-10m+5n-mn+8 is simplest form

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3 years ago

What is the area of the shaded region ?

aleksklad [387]

25 this is it I don’t know if u understand it because I don’t lol

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3 years ago

What is 65 miles in 2.7 gallons unit rate rounded to the nearest hundredth

STALIN [3.7K]

You can't convert miles into gallons those are two different units of measurement.

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3 years ago

Read 2 more answers

Can someone help me find the equivalent expressions to the picture below? I’m having trouble

miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is $\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$ .

Now we will solve this expression with the help of law of exponents.

$\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}$

$=2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}$

$=2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}$

$=2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}$

$=2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }$ [Option 2]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$ [Option 1]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$

$=(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }$

$=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }$ [Option 3]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }$

$=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}$ [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

6 0

2 years ago