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alina1380 [7]
3 years ago
13

F(x)=-5x+2 and g(x)=1/2x+4 find f(g(12))

Mathematics
1 answer:
skad [1K]3 years ago
3 0

Answer:

- 48

Step-by-step explanation:

If you want to learn more about this concept, it's called composition of functions.

First, you plug in g(x) as if it were x into f(x).

f(g(x))= -5 (1/2x + 4) + 2

= -5/2x - 20 + 2

= -5/2x - 18

Then, plug in the value given, as x.

= -5/2 (12) - 18

= -30 - 18

= - 48

I hope this helped!

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Find the base of an isosceles triangle whose area is 60 cm² and the length of one of its equal sides is 13 cm. ​
Kobotan [32]

Answer:

Base = 24 cm or 10cm

Step-by-step explanation:

REMEMBER:
An isosceles triangle ABC with base BC = ‘b' & height AD = ‘h' & its equal sides =13 cm & area = 60 cm²

Using the formulas

A=\frac{bh_b}{2}

h_b=\sqrt{a^2-\frac{b^2}{4} }

There are 2 solutions for b

b=2\sqrt{2}   \frac{A}{\sqrt{a^2+\sqrt{a^4-4A^2} } } =2*\sqrt{2} *\frac{60}{\sqrt{13^2+\sqrt{13^4-4*60^2} } } ≈ 10cmb=2\sqrt{2}   \frac{A}{\sqrt{a^2+\sqrt{a^4-4A^2} } } =2*\sqrt{2} *\frac{60}{\sqrt{13^2-\sqrt{13^4-4*60^2} } }=24cm

Less complex:

Area of a triangle = 1/2 * b * h = 60

=> h = 120/b

In right triangle ABD

13² = h² + b² /4 ( by Pythagoras law)

=>169 = 120²/b² + b²/4

=>676 b² = 57600 + b^4

=> b^4 - 676 b² + 57600 = 0

=> b² = 676 +- √(676² - 4*57600) / 2

=> b²= 676 +- √(226576) /2

=> b² = (676 +- 476 )/2

=> b² = 1152/2 , 200 /2

=> b² = 576 , 100

=> b = 24, 10

So, Base = 24 cm or 10cm

4 0
2 years ago
6 + x is an example of _____. a formula a constant a variable an expression
PtichkaEL [24]

Answer:

it is an example of an expression

Step-by-step explanation:

it is an example of an expression because it is asking a question without an equal sign. so its not a question, but an expression

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Answer:

C.(3|-4)

Step-by-step explanation:

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For positive test result, the number of those who did not lie and lie are 11 and 41, respectively. Those of the negative test re
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P(subject lied | negative results) = 4/19

P(negative results | subject lied) = 8/49

 

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

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