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3 years ago

(x4 − 9x + 5x7) + (5x − 10 + 3x4 − 2x2)

Mathematics

1 answer:

Svetlanka [38]3 years ago

4 0

Answer:

(x^4 - 9·x + 5·x^7) + (5·x - 10 + 3·x^4 - 2·x^2)

x^4 - 9·x + 5·x^7 + 5·x - 10 + 3·x^4 - 2·x^2

5·x^7 + x^4 + 3·x^4 - 2·x^2 - 9·x + 5·x - 10

5·x^7 + 4·x^4 - 2·x^2 - 4·x - 10

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3 years ago

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Which of the following have the property that a(x)=a−1(x)? I. y=x II. y=1/x III.y=x^2 IV. y=x^3 A. I and II, only B. IV, only C.

valentina_108 [34]

Answer:

Correct answer:

A. I and II

Step-by-step explanation:

First of all, let us have a look at the steps of finding inverse of a function.

1. Replace y with x and x with y.

2. Solve for y.

3. Replace y with $f^{-1}(x)$

Given that:

$I.\ y=x \\II.\ y=\dfrac{1}x \\III.\ y=x^2 \\IV.\ y=x^3$

Now, let us find inverse of each option one by one.

I. y = x, a(x) = x

Replacing y with and x with y:

x = y

x = $a^{-1}(x)$ = $a(x)$ Hence, I is true.

II. $y =\dfrac{1}{x}$

Replacing y with and x with y:

$x =\dfrac{1}{y}$

$x=\dfrac{1}{a^{-1}(x)}$

$\Rightarrow a^{-1}(x) = \dfrac{1}{x}$

$a^{-1}(x)$ = $a(x)$ Hence, II is true.

III. $y =x^{2}$

Replacing y with and x with y:

$x =y^{2}\\\Rightarrow y = \sqrt x\\\Rightarrow a^{-1}(x) = \sqrt{x} \ne a(x)$

Hence, III is not true.

IV. $y =x^{3}$

Replacing y with and x with y:

$x =y^{3}\\\Rightarrow y = \sqrt[3] x\\\Rightarrow a^{-1}(x) = \sqrt[3]{x} \ne a(x)$

Hence, IV is not true.

Correct answer:

A. I and II

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3 years ago

A=1/2h(b+c) solve for h

pogonyaev

1) Multiply both sides of the equation by 2:

2(A)=2[1/2h(b+c)
2A=2/2h(b+c)
2A=1h(b+c)
2A=h(b+c)

2) Divide both sides of the equation by (b+c):
(2A)/(b+c)=[h(b+c)]/(b+c)
2A/(b+c)=h
h=2A/(b+c)

Answer: $h=2 \frac{A}{(b+c)}$

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Answer:

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Step-by-step explanation:

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3 years ago

Find the value of x. Round the length to the nearest tenth. The diagram is not drawn to scale.

Gekata [30.6K]

Answer:

x ≈ 2879.4 m

Step-by-step explanation:

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Measure of ∠DAB = 90°

Measure of ∠BAC = 90° - 10° = 80°

By applying cosine rule in ΔABC,

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