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4 years ago

If ybx+ 3 then does 4 = cdb ?

Mathematics

2 answers:

olga nikolaevna [1]4 years ago

7 0

Answer:

now ion kno what type of math this is btfu but

no 4 should equal dbc which is anwer 4

hope i helped

son4ous [18]4 years ago

5 0

Answer:

nahh slime it should be option 4 bdc :)- $latt

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Given 2 f (x) = -5x^2 – 8x + 3 and g(x) = -3x^2 + 7x + 11, find (f +g) (x)

denis23 [38]

Answer:

$\boxed{\sf \ \ \ (f+g)(x)=-8x^2-x+14 \ \ \ }$

Step-by-step explanation:

Hello,

$f(x)=-5x^2-8x+3 \ and\\g(x)=-3x^2+7x+11 \ so\\(f+g)(x)=f(x)+g(x)=-5x^2-8x+3-3x^2+7x+11\\=-8x^2-x+14$

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3 years ago

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3 years ago

tanisha office is on the 9 floor of an office building . She parks her car in the underground parking garage 4 floors below grou

Leni [432]

Answer: 13

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4 0

4 years ago

3+5i/-2+3i perform operation

melisa1 [442]

Answer:

$\dfrac{9}{13}-\dfrac{19}{13}i$

Step-by-step explanation:

Remember $i^2=-1$

You are given the fraction $\dfrac{3+5i}{-2+3i}$

First, multiply the numerator and the denominator by -2-3i:

$\dfrac{3+5i}{-2+3i}=\dfrac{(3+5i)(-2-3i)}{(-2+3i)(-2-3i)}=\dfrac{(3+5i)(-2-3i)}{(-2)^2-(3i)^2}=\dfrac{(3+5i)(-2-3i)}{4-9i^2}=\dfrac{(3+5i)(-2-3i)}{4+9}$

This gives you 13 in denominator, now multiply two complex numbers in numerator:

$(3+5i)(-2-3i)=-6-9i-10i-15i^2=-6-19i+15=9-19i$

Thus, the initial fraction is

$\dfrac{9-19i}{13}=\dfrac{9}{13}-\dfrac{19}{13}i$

4 0

3 years ago

How do I factor e^2-10e+25

olga nikolaevna [1]

Answer: (e-5)^2

Step-by-step explanation:

The box method is a good way to solve this, but for a quick explanation, start by finding a single value that when added gets -10 and when multiplied gets +25. We can find that this term is -5. We the can situate -5 into (e-5)^2 equation, where if you FOIL, you will be back at the unfactored equation. Hope this helps

7 0

3 years ago