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Katarina [22]
3 years ago
6

URGET!! HAVE 3 MINS LEFT please help me and will mark brainliest!!!!

Mathematics
2 answers:
tensa zangetsu [6.8K]3 years ago
4 0

Answer:

1.) A, B, and E

2.) (0, 0)

3.) -2

4.) f(x) ≥ 3

5.) -7

Step-by-step explanation:

valentinak56 [21]3 years ago
3 0

Answer:

A, B, and E

(0, 0)

-2

 f(x) ≥ 3

 -7

I hope this helps I LUV YA

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3 years ago
PLS HELP ASAP WILL GIVE BRAINLIEST PLSSSS
Dmitrij [34]

Answer:

I won't write it in paragraph but use this:

  1. angle A + angle B + angle C = 180 so when one angle is missing add the other 2 and subtract it to 180, that would be the answer for the missing angle; applies only to triangle (all kinds)
  2. if its the exterior angle, add them all like the one above but subtract it instead to any shapes or if it is adjacent to a given degree, you can always subtract it to 180 to get the exterior
  3. if its a different shape, count the sides and subtract it by two then multiply by 180; thats the degree for the total angles of the shape
  4. if its variable, do the third one (unless you know what it is) add them all like the first one and subtract to the answer in the #3
6 0
3 years ago
If x = 2 and y = -3, what is the value of 2x^2−3xy−2y^2<br> HELPPP
oksano4ka [1.4K]

Answer:

8

Step-by-step explanation:

2(2)^2‒3(2*‒3)‒2(‒3)^2

2*4‒3(‒6)‒2(9)

8‒(‒18)‒18

8+18‒18

26‒18

8

i think it is easy to understand &i hope it will helps you

8 0
3 years ago
Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}
miskamm [114]

Answer:

-3+i\sqrt{3} , 1+\sqrt{3}

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

\alpha = x+iy\\\beta = x-iy

since they are conjugates

\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}

\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}

Imaginary part of the above =0

i.e. \sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1

So the value of alpha = -3+i\sqrt{3} , 1+\sqrt{3}

3 0
3 years ago
What is 69.6 × 0.85​
Vladimir [108]

Answer:59.16

Step-by-step explanation:

Calculator

6 0
4 years ago
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