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lozanna [386]
3 years ago
9

What is the variable to y = 64 (18)

Mathematics
2 answers:
Sladkaya [172]3 years ago
7 0

Answer:

y=1152

Step-by-step explanation:

PolarNik [594]3 years ago
7 0

Answer:

the varible equals 1152

Step-by-step explanation:

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(X-2)^2 is subtracted from 3x^2-4x
Andrew [12]

Answer:

-2x^2-8x+4

Step-by-step explanation:

4 0
3 years ago
Is 3/2 an allowable probability? Explain.
nadezda [96]
No, probably is on a scale between 0 and 1. Anything beyond those numbers is mathematically impossible. Since 3/2 = 1.5, it is not valid
7 0
3 years ago
For each logarithmic equation, write an equivalent equation in exponential form
Montano1993 [528]

Answer:

  • loga b = c             ⇔   a^c = b

<u>Apply the property above:</u>

  • 1.  ln 618 = p   ⇔ e^p = 618
  • 2. ln q = 2       ⇔ e^2 = q
  • 3. ln 100 = t     ⇔  e^t = 100
  • 4. ln (e^3) = 3  ⇔  e^3 = e^3
8 0
2 years ago
On a coordinate plane, 2 triangles are shown. The first triangle has points S (3, 5), R (5, 0), and Q (1, 0). The second triangl
enyata [817]

Answer:

<u>B. (x, y) → (–x, –y).</u>

Step-by-step explanation: This is the correct answer on <u>Edge 2021</u>, just did the assignment. Hope this helps ^-^.

6 0
2 years ago
What are the solutions to the following system?
bija089 [108]

The solutions are (\sqrt{2},-1) \text { and }(-\sqrt{2},-1).

Solution:

Given equation:

-2 x^{2}+y=-5

Add 2 x^{2} on both sides.

-2 x^{2}+y+2 x^{2}=-5+2 x^{2}

y=-5+2 x^{2} -------- (1)

y=-3 x^{2}+5 (given)  -------- (2)

Equate (1) and (2).

-5+2 x^{2}=-3 x^{2}+5

Add 3 x^{2} on both sides.

-5+2 x^{2}+3 x^{2}=-3 x^{2}+5 +3 x^{2}

-5+5x^{2}=5

Add 5 on both sides.

-5+5x^{2}+5=5+5

5x^{2}=10

Divide by 5 on both sides, we get

x^{2}=2

Taking square root on both sides, we get

x=\sqrt{2}, x=-\sqrt{2}

Substitute x=\sqrt{2} in (1).

y=-5+2(\sqrt{2})^2

y=-5+4

y=-1

Substitute x=-\sqrt{2} in (1).

y=-5+2(-\sqrt{2})^2

y=-5+4

y=-1

Therefore the solutions are (\sqrt{2},-1) \text { and }(-\sqrt{2},-1).

Option C is the correct answer.

6 0
3 years ago
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