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Strike441 [17]
4 years ago
14

12 + 9/3 exponent 2 What is the answer?

Mathematics
2 answers:
UkoKoshka [18]4 years ago
4 0
I think there can be two answers.

If the 9/3 is not a fraction:                          If it's a fraction:

12 + 9/3²                                                    12 + 9/3²
12 + 9/9                                                     12  +  9  =                                           
12+1 =                                                        21
13
ira [324]4 years ago
3 0
9/3 = 3
3 exponent 3 = 27.

27 + 12 = 39
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2/3 + n + 6 = 3/4<br>what is n with an explanation​
Talja [164]

Answer: -6

Step-by-step explanation:

2/3 + n + 6 = 3/4

-2/3              -2/3

n + 6 = 1/12

   -6      -6

n = - 6

5 0
3 years ago
Which of the following cannot be used to prove
Ksju [112]

Given:

Different types of congruence postulates.

To find:

Which cannot be used to prove  that two triangles are congruent?

Solution:

According to AAS congruence postulate, if two angles and a non including sides of two triangles are congruent, then triangles are congruent.

According to SAS congruence postulate, if two sides and an including angle of two triangles are congruent, then triangles are congruent.

According to SSS congruence postulate, if all three sides of two triangles are congruent, then triangles are congruent.

AAA states that all three angles of two triangles are equal and no information about sides.

So, it is a similarity postulate not congruent postulate. According to AAA two triangles are similar not congruent.

Therefore, the correct option is D.

4 0
3 years ago
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Find the perimeter of ADE
irga5000 [103]
At D point in the bottom then up and to the right upper corner to Letter B
5 0
3 years ago
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Mamont248 [21]

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5 0
3 years ago
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Write the equation of the conic section below.
tangare [24]

Answer:

(x-3)^2=8(y+1)

Step-by-step explanation:

Standard form of a parabola with a vertical axis of symmetry:

(x-h)^2=4p(y-k) \quad \textsf{where}\:p\neq 0

\textsf{Vertex}=(h, k)

\textsf{Focus}=(h,k+p)

\textsf{Directrix}:y=(k-p)

\textsf{Axis of symmetry}:h=k

If p > 0, the parabola opens upwards, and if p < 0, the parabola opens downwards.

From inspection of the graph:

  • Vertex = (3, -1)
  • Directrix = y = -3

Therefore:

  • h = 3
  • k = -1
  • k - p = -3

Use the Directrix equation to find p

⇒ y = (k - p)

⇒ -3 = -1 - p

⇒ p = 2

Therefore the equation of the conic section is:

\implies (x-3)^2=4\cdot 2(y-(-1))

\implies (x-3)^2=8(y+1)

Rearranging in standard form ax^2+bx+c:

\implies (x-3)^2=8y+8

\implies x^2-6x+9-8=8y

\implies x^2-6x+1=8y

\implies y=\dfrac{1}{8}x^2-\dfrac{3}{4}x+\dfrac{1}{8}

7 0
2 years ago
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