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4 years ago

9

Select the correct answer from the drop-down menu. Find the missing term. The sum of 2ab2 and (-5ab2) is the same as the sum of

(-6ab2) and -2ab^2 2ab^2 -3ab^2 3ab^2 .

1 answer:

AysviL [449]4 years ago

3 0

<span>Let x be a missing term

(-6ab</span>²) + x = 2ab² + (-5ab²)
-6ab² + x = 2ab² - 5ab²
x = 2ab² - 5ab² + 6ab²
x = 3ab²

Answer: 3ab²

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ad-work [718]

The answer is:
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3 years ago

9. Find the Highest Common Factor of

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C. 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60

1,000: 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000

Now we find the common numbers. One doesn’t count as when multiplied later on, it will not change anything.

60: 2, 4, 5, 10, 20

1,000: 2, 4, 5, 10, 20

The highest common factor is 20 because it’s, well, the highest number.

D. Do the same thing for D.

24: 1, 2, 3, 4, 6, 8, 12, 24

880: 1, 2, 4, 5, 8, 10, 11, 16, 20, 22, 40, 44, 55, 80, 88, 110, 176, 220, 440, 880

20 and 880: 2, 4, 8

8 is the Highest Common Factor.

E. Do the same thing with E.

90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

1,000: 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000

90 and 1000: 2, 5, 10

10 is the Highest Common Factor.

3 0

3 years ago

Read 2 more answers

Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes

azamat

<h3>Answer:</h3>

Equation of the ellipse = 3x² + 5y² = 32

<h3>Step-by-step explanation:</h3>

<h2>Given:</h2>

The centre of the ellipse is at the origin and the X axis is the major axis

It passes through the points (-3, 1) and (2, -2)

<h2>To Find:</h2>

The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

$\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1$

Given that the ellipse passes through the point (-3, 1)

Hence,

$\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1$

Cross multiplying we get,

9b² + a² = 1 ²× a²b²
a²b² = 9b² + a²

Multiply by 4 on both sides,

4a²b² = 36b² + 4a²------(1)

Also by given the ellipse passes through the point (2, -2)

Substituting this,

$\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1$

Cross multiply,

4b² + 4a² = 1 × a²b²
a²b² = 4b² + 4a²-------(2)

Subtracting equations 2 and 1,

3a²b² = 32b²
3a² = 32
a² = 32/3----(3)

Substituting in 2,

32/3 × b² = 4b² + 4 × 32/3
32/3 b² = 4b² + 128/3
32/3 b² = (12b² + 128)/3
32b² = 12b² + 128
20b² = 128
b² = 128/20 = 32/5

Substituting the values in the equation for ellipse,

$\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1$

$\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1$

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>

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3 years ago

What is √117 simplified

kenny6666 [7]

Answer:

10.82

Step-by-step explanation:

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3 years ago

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Jordan throws a ball from a cliff 12 feet above the ground with an initial velocity of 32 feet per second. In the following func

lbvjy [14]

Answer:

Your answer will be the third function

Step-by-step explanation:

The base function you need to know is h(t)= 1/2at^2

Your acceleration in this problem is going to be gravity which they give to you, 32 feet per second squared. Since the ball is falling, it means it will have negative acceleration. Now you have the equation h(t)= -16t^2. The final step is to add the initial height from which the ball was dropped giving you: h(t)= -16t^2 +12

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