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ozzi
4 years ago
8

Whitch expression is equivalent to p + (q + r )

Mathematics
1 answer:
fredd [130]4 years ago
4 0
The answer will be p+q+r, or p+r+q, or q+p+r. Hope it help!
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First I broke the numbers apart first I broke 3.54 + 3.00 + 0.50 + 0.04 + 12.09+ into 12. 00+ 0.09
vovikov84 [41]
Avocados from Mexico
3 0
3 years ago
What is the greatest common factor of 9x^2<br> 2<br> and 6x?
zheka24 [161]

Answer:

The greatest common factor is 1.

Step-by-step explanation:

In this question we will factorize the expressions given and select the common factors among all.

For 9x² = 1×3×3×X×X

For 2 = 1×2

For 6x = 1×2×3×X

So we find the common factor in all expressions is = 1

Answer is 1.

7 0
4 years ago
Help ASAP please ...
cupoosta [38]

Answer:

Angle Q

Step-by-step explanation:

As I said on my previous posts, all you need to remember is CPCTC (corresponding parts of congruent triangles are congruent)

Q corresponds to B, and is thus congruent

8 0
3 years ago
Read 2 more answers
Find the solutions for a triangle with a = 16, c =12, and B = 63º.
maks197457 [2]

Answer:

b. A = 71.6°; C = 45.40°; b =15.0

Step-by-step explanation:

The missing values can be found with the help of the Law of Cosine and properties of triangles:

Side b (Law of Cosine)

b = \sqrt{a^{2}+c^{2}-2\cdot a \cdot c \cdot \cos B}

b = \sqrt{16^{2}+12^{2}-2\cdot (16)\cdot (12) \cdot \cos 63^{\circ}}

b \approx 15.022

Angle A (Law of Cosine)

\cos A = -\frac{a^{2} - b^{2}-c^{2}}{2\cdot b \cdot c}

\cos A = - \frac{16^{2}-15.022^{2}-12^{2}}{2\cdot (15.022)\cdot (12)}

\cos A = 0.315

A= \cos^{-1} 0.315

A \approx 71.639^{\circ}

Angle C (Sum of internal angles in triangles)

C = 180^{\circ} - 63^{\circ} - 71.639^{\circ}

C = 45.361^{\circ}

Hence, the right answer is B.

6 0
3 years ago
Solve this inequality algebracially. |3-x|&gt;10
xeze [42]

Answer:

x < -7 or x > 13

Step-by-step explanation:

3 - x < -10 or 3 - x > 10

-x < -13      or    -x > 7

x > 13         or    x < -7

Answer: x < -7 or x > 13

8 0
2 years ago
Read 2 more answers
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