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3 years ago

Suppose f(x) = x2 and g(x) = 2x − 3. What is the value of g(4) + f(−3)?

Mathematics

1 answer:

MAVERICK [17]3 years ago

8 0

In order to find the value of g(4)+f(-3), we must substitute those values into the functions first.

Substitute -3 into f(x).

f(x)=x²

f(-3)=(-3)²

f(-3)=9

Substitute 4 into g(x).

g(x)=2x-3

g(4)=2(4)-3

Using the rules of PEMDAS, multiply first then subtract.

g(4)=8-3

g(4)=5

Therefore, if f(-3) is 9 and g(4) is 5, we must add the two values together.

5+9=14

The value of g(4)+f(-3) is 14.

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3 years ago

X^2 -50x+456=0 solve this algebraiclly find x

bagirrra123 [75]

Answer:

$x=-406$

Step-by-step explanation:

$x^2-50x+456=0$

$x^2-50x=-456$

This is because we subtract 456 from each side.

$x-50=-456$

We then divide the whole equation by x .

$x=-406$

Adding 50 to both sides gives us x by itself, hope this helps :D

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Given: ∆ABC –iso. ∆, m∠BAC = 120°

lidiya [134]

Answer: $P\triangle ADH = (a+b+\sqrt{a^2+b^2} )$ unit

Step-by-step explanation:

Since Here Δ ABC is a isosceles triangle.

Also, $AH\perp BC$ and $HD\perp AC$

Thus, ∠ ADH = 90°

That is, Δ ADH is the right angle triangle.

In which, AD = a unit and HD = b unit

And, By the definition of Pythagoras theorem,

$AH^2 = HD^2 + AD^2$

⇒ $AH^2 = a^2 +b^2$

⇒ $AH = \sqrt{a^2+b^2}$

Since, the perimeter of a triangle = sum of the all sides of the triangle.

Therefore, perimeter of ΔADH = AD + DH + AH

= $a + b + \sqrt{a^2+b^2}$ unit

4 0

3 years ago