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Ilia_Sergeevich [38]
3 years ago
7

Someone help please!!

Mathematics
1 answer:
mash [69]3 years ago
8 0

Answer:

y = 6 - 2x

6 - 2 X 0 = 6

6 - 2 X 1 = 4

6 - 2 X 2 = 2

6 - 2 X 3 = 0

6 - 2 X 4 = -2

6 - 2 X 5 = -4

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3 years ago
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Brian scored 120 on his test the first time he took the test. The second time he took the test,
Brums [2.3K]

Answer:

b. 25%

Step-by-step explanation:

(new - old)/ old * 100%

150 - 120 = 30 ... 30/120 = .25 * 100% = 25%

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2 years ago
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Which of the following expressions is equal to -4x^2-36
liubo4ka [24]

Answer:

Step-by-step explanation:

-4x^2-36 can be rewritten as -4(x^2 + 9).  The quantity in parentheses does not have real roots, but does have the imaginary roots ±3i.

Next time, please share the possible answer choices.

6 0
3 years ago
PLEASE HELP AND SHOW WORK IF YOU CAN​
Murljashka [212]

Answer:

c) XY > RS

= true

Step-by-step explanation:

Firstly know the meaning of each sign given

> means <u>is greater than</u>

= means <u>is equal to</u>

a)XZ > ST

XZ = 8

ST = 11 - greater

= false

b) XY = RS

Due to the angles presented, we conclude that the greater the angle, the longer the unmeasured line.

XY is longer than RS

= false

c) XY > RS

= true

d) RS > XY

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3 0
3 years ago
Find the standard form of the equation of the hyperbola satisfying the given conditions: X intercept +/- 6; foci at (-10,0) and
uysha [10]

Answer:

\frac{x^{2}}{36} - \frac{y^{2}}{64}=1

Step-by-step explanation:

Given an hyperbola with the following conditions:

  • Foci at (-10,0) and (10,0)
  • x-intercept +/- 6;

The following holds:

  • The center is midway between the foci, so the center must be at (h, k) = (0, 0).
  • The foci are 10 units to either side of the center, so c = 10 and c^2 = 100
  • The center lies on the origin, so the two x-intercepts must then also be the hyperbola's vertices.

Since the intercepts are 6 units to either side of the center, then a = 6 and a^2 = 36.

Then, a^2+b^2=c^2\\b^2=100-36=64

Therefore, substituting a^2 = 36. and b^2=64 into the standard form

\frac{x^{2}}{a^2} - \frac{y^{2}}{b^2}=1\\We \: have:\\ \dfrac{x^{2}}{36} - \dfrac{y^{2}}{64}=1

4 0
3 years ago
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