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3 years ago

6

Please answer this question...No spam

1 answer:

mariarad [96]3 years ago

8 0

Answer:

(d) (x² + y² + z²)³ = 27x²y²z²

Step-by-step explanation:

<u>Use the identity:</u>

a³ + b³ + c³ = 3abc if a + b + c = 0

<u>Substitute as follows:</u>

a = $x^{2/3}$ , b = $y^{2/3}$ , c = $z^{2/3}$

<u>The sum of the cubes is:</u>

x² + y² + z² = 3(xyz) $^{2/3}$

<u>Cube both sides:</u>

(x² + y² + z²)³ = 27x²y²z²

Correct choice is (d)

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A rectangle is removed from the right hand corner of a square board. What is the area of the remaining board in square feet?

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4 years ago

A rectangle has a length of 4 feet and a perimeter of 14 feet. What is the perimeter of a similar rectangle with a width of 9 fe

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Answer:

Option C. $42\ ft$

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is equal to the scale factor and the ratio of its perimeter is equal to the scale factor too

step 1

Find the width of the rectangle that has a length of 4 feet and a perimeter of 14 feet

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substitute the values and solve for W

$14=2(4+W)$

$W=7-4=3\ ft$

step 2

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To find the scale factor divide the width of the larger rectangle by the width of the smaller rectangle

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3 years ago

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3 years ago

Indicate the sum of the terms in each of the following examples, and simplify the result by combining like terms:

katrin [286]

Answer:

The sum of the terms is equal to $5r^2-14r+3s$ .

The value of the sum at $r=1,s=3$ is 0.

Step-by-step explanation:

Given:

The terms that are given are:

$10r^2, -6r, +5s, -8r, +2s, -5r^2, -4s$

The sum of the terms is the addition of the given terms and is given as:

$=10r^2+ (-6r)+5s+ (-8r)+2s + (-5r^2) + (-4s)$

$=10r^2-6r+5s-8r+2s-5r^2-4s$

Now, combining like terms. Like terms means that are of the same type are grouped together. There are 3 different terms here 'r²', 'r' and 's'.

The terms containing 'r²' are: $10r^2, -5r^2$

The terms containing 'r' are: $-8r, -6r$

The terms containing 's' are: $5s, 2s, -4s$

Now, combining the like terms, we get:

$=(10r^2-5r^2)+(-8r-6r)+(5s + 2s -4s)$

$=(10-5)r^2+(-8-6)r+(5+2-4)s$

$=5r^2+(-14)r+3s$

$=5r^2-14r+3s$

Therefore, the final expression is equal to:

$=5r^2-14r+3s$

Now, plug in $r=1,s=3$ . This gives,

$=5(1)^2-14(1)+3(3)$

$=5\times 1-14+9$

$=5-14+9$

$=0$

Therefore, the value of the sum of all the terms at $r=1,s=3$ is 0.

4 0

3 years ago