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Lubov Fominskaja [6]
3 years ago
6

Please answer, will give brainliest.

Mathematics
2 answers:
vladimir1956 [14]3 years ago
6 0

Answer:

third one

Step-by-step explanation:

Paul [167]3 years ago
5 0

Answer:

I think one solution

Step-by-step explanation:

*Sorry if this did not help. I'm only in 5th grade*

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PLS I need help, this is due today. thank u.
mariarad [96]
Here, you would complete it with the Pythagorean Theorem.

So, A^2 + B^2 = C^2

So 14^2 + B^2 = 14^2

This equals = B = 11.49
6 0
3 years ago
Read 2 more answers
PLEASEEE HELPPPP ASAPPPP
stellarik [79]

\cos 27^{\circ} = \dfrac{KL}x\\\\\implies \cos 27^{\circ} = \dfrac{6.3}x\\\\\implies x = \dfrac{6.3}{\cos 27^{\circ}}=7.1~ \text{units.}

6 0
2 years ago
4TH TIME ASKING THIS!!! Please help me! Someone pleaseeee. I need the correct answers. I don’t want to fail
Alexeev081 [22]

Answer:

The functions are inverses; f(g(x)) = x ⇒ answer D

h^{-1}(x)=\sqrt{\frac{x+1}{3}} ⇒ answer D

Step-by-step explanation:

* <em>Lets explain how to find the inverse of a function</em>

- Let f(x) = y

- Exchange x and y

- Solve to find the new y

- The new y = f^{-1}(x)

* <em>Lets use these steps to solve the problems</em>

∵ f(x)=\sqrt{x-3}

∵ f(x) = y

∴ y=\sqrt{x-3}

- Exchange x and y

∴ x=\sqrt{y-3}

- Square the two sides

∴ x² = y - 3

- Add 3 to both sides

∴ x² + 3 = y

- Change y by f^{-1}(x)

∴ f^{-1}(x)=x^{2}+3

∵ g(x) = x² + 3

∴ f^{-1}(x)=g(x)

∴ <u><em>The functions are inverses to each other</em></u>

* <em>Now lets find f(g(x))</em>

- To find f(g(x)) substitute x in f(x) by g(x)

∵ f(x)=\sqrt{x-3}

∵ g(x) = x² + 3

∴ f(g(x))=\sqrt{(x^{2}+3)-3}=\sqrt{x^{2}+3-3}=\sqrt{x^{2}}=x

∴ <u><em>f(g(x)) = x</em></u>

∴ The functions are inverses; f(g(x)) = x

* <em>Lets find the inverse of h(x)</em>

∵ h(x) = 3x² - 1 where x ≥ 0

- Let h(x) = y

∴ y = 3x² - 1

- Exchange x and y

∴ x = 3y² - 1

- Add 1 to both sides

∴ x + 1 = 3y²

- Divide both sides by 3

∴ \frac{x + 1}{3}=y^{2}

- Take √ for both sides

∴ ± \sqrt{\frac{x+1}{3}}=y

∵ x ≥ 0

∴ We will chose the positive value of the square root

∴ \sqrt{\frac{x+1}{3}}=y

- replace y by h^{-1}(x)

∴ h^{-1}(x)=\sqrt{\frac{x+1}{3}}

4 0
3 years ago
I dont get how to solve this?
Alex

Answer:

Step-by-step explanation:

In standard form this could be rewritten as | b - 0 | > 6, which in words, says,

"b is greater than 6 units away from 0 on a number line".  If b is greater than 6 units away, both positive 6 and negative 6 are 6 units away from 0.  Because the values we need are greater than 6 units away from 0, we are starting at both 6 and -6 and moving away from 0.  

The "greater than" symbol indicates that this is a disjunction so the word "or" is used as opposed to in a conjunction where the word "and" is used.

Therefore, our solutions are

b < -6 or b > 6, the 3rd choice down

5 0
3 years ago
Pls help!! I need the answer immediately
Strike441 [17]

Answer:

i). x³ + 9x² + yz - 15

ii). -21m³np - 8p⁵q + mnp + 4mn + 100

Step-by-step explanation:

Question (38)

i). Two expressions are -5x² - 4yz + 15 and x³+ 4x²- 3yz

  By subtracting expression (1) from expression (2) we can the expression by addition which we can get expression (1).

 (x³+ 4x²- 3yz) - (-5x² - 4yz + 15) = x³ + 4x² - 3yz + 5x² + 4yz - 15

                                                    = x³ + 9x² + yz - 15

ii). -15m³np + 2p⁵q - 6m³pn + mnp + 4mn - 10qp⁵+ 100

  = (-15m³np - 6m³np) + (2p⁵q - 10qp⁵) + mnp + 4mn + 100

  = -21m³np - 8p⁵q + mnp + 4mn + 100

4 0
3 years ago
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