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sleet_krkn [62]
3 years ago
14

F(x) = x^2 - 3x+ 5 g(x) = 2x^2 - 4x - 11 what is h(x) = f(x) + g(x)

Mathematics
2 answers:
sashaice [31]3 years ago
6 0

Answer:

\huge\boxed{(f+g)(x)=3x^2-7x-6}

Step-by-step explanation:

f(x)=x^2-3x+5\\\\g(x)=2x^2-4x-11\\\\(f+g)(x)=f(x)+g(x)\\\\\text{substitute}\\\\(f+g)(x)=(x^2-3x+5)+(2x^2-4x-11)\\\\(f+g)(x)=x^2-3x+5+2x^2-4x-11\\\\\text{combine like terms}\\\\(f+g)(x)=(x^2+2x^2)+(-3x-4x)+(5-11)\\\\(f+g)(x)=3x^2-7x-6

iogann1982 [59]3 years ago
5 0

Hey there! :)

Answer:

h(x) = 3x² - 7x - 6

Step-by-step explanation:

Calculate h(x) by adding the two polynomials:

h(x) = f(x) + g(x):

h(x) = x² - 3x + 5 + 2x² - 4x - 11

Combine like terms:

h(x) = 3x² - 7x - 6

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3 0
3 years ago
Read 2 more answers
JK=2X KL=X+2 and JK=5X-10
elixir [45]

JK + KL = JL

JK = 2x, KL = x + 2, JL = 5x - 10

therefore we heve the equation:

2x + (x + 2) = 5x - 10

3x + 2 = 5x - 10 |-2

3x = 5x - 12 |-5x

-2x = -12 |:(-2)

x = 6

JK = 2x → JK = 2(6) = 12

KL = x + 2 → KL = 6 + 2 = 8

JL = 5x - 10 → JL = 5(6) - 10 = 30 - 10 = 20

Check:

JK + KL = JL

12 + 8 = 20 CORRECT :)

5 0
3 years ago
At 6:00 A.M. the temperature was 33°F. By noon the temperature had increased by 10°F and by 3:00 P.M. it had increased another 1
chubhunter [2.5K]

So let's see here...


6:00 A.M. temperature is 33°F.

12:00 P.M temperature increased by 10°F making it 43°F.

3:00 P.M. temperature increased by another 12°F making it 55°F.

At 10:00 P.M it would decrease by 15°F making it 40°F.

The temperature would need to fall (or decrease) by 7°F to reach the original temperature of 33°F

Hope this helped!


3 0
3 years ago
Alguém sabe como faz essas duas questões ?
joja [24]

Minha resposta: não.

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Nimfa-mama [501]

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