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3 years ago

I'm stuck on two polynomial questions and any help would be greatly appreciated!

Mathematics

1 answer:

CaHeK987 [17]3 years ago

4 0

Answer:

1) 25
2) 84

Step-by-step explanation:

a²x² - b²y² = (ax + by)(ax - by)

Since (4x - 3y) is one of the factors, we have:

a = 4, b = ± 3

<u>The value of a² + b² is:</u>

4² + (±3)² = 16 + 9 = 25

x² + 19x + c, has a factor of (x + 7).

<u>It means the equation:</u>

x² + 19x + c = 0 has one root of x₁ = -7.

<u>Sum of the roots:</u>

x₂ + (- 7) = -19/1
x₂ = -19 + 7
x₂ = -12

<u>Product of the roots:</u>

x₂*(-7) = c
-12*(-7) = c
c = 84

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Step-by-step explanation:

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3 years ago

The scale on a map is 1 :50000.

mestny [16]

Answer:

The actual area of the field in square kilometers is $30000000 km^2$

Step-by-step explanation:

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3 years ago

What is the area of this compound shape?!

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3 years ago

Read 2 more answers

Two friends, Ryan and Tressa, are on vacation. They decide to go bungee jumping off a platform. Their height from the ground can

eimsori [14]

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Step-by-step explanation:

Step 1 :

Given,

The function representing the height is y = 4x^2-32x+80

where the x represents the time after they jumped and y represents their height above ground.

We need to find the platform's height.

Step 2 :

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Step 3:

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4 years ago

Which equation justifies why seven to the one third power equals the cube root of seven?

Rudiy27

The equation that justifies why seven to the one third power equals the cube root of seven is seven to the one third power all raised to the third power equals seven to the one third times three power equals seven

<h3>How to solve an equation?</h3>

using indices law,

$\sqrt[n]{x} = x^{1 / n}$

Therefore, let's justify

$7^{1/3} = \sqrt[3]{7}$

Hence, the justification is as follows;

$(7^{1 /3}) ^{3} =(7^{1/3})^{3} = 7$

Therefore, the justification of is that seven to the one third power all raised to the third power equals seven to the one third times three power equals seven

learn more on equation here: brainly.com/question/13426221

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2 years ago