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kobusy [5.1K]
3 years ago
5

Evaluate x(+3) (3+y). for x = 2, y = 5, and z = 1. pls help

Mathematics
2 answers:
mart [117]3 years ago
7 0

Answer

The answer is most likely 2 (or C).

Why?

The top of the question simplified into number is 2(5 + 3), which equals 16.

The bottom is (3 +5)1, which equals 8.

Now divide 16/8, which equals 2, and there's your answer.

(Hope this helps you! ^^)

pentagon [3]3 years ago
3 0

Answer:

2

Step-by-step explanation:

First, substitute the x’s and y’s:

2(5+3)/(3+5)1

since the one doesn’t make a difference, we don’t need to write it in.

2(5+3)/(3+5)

next, solve the parenthesis first:

5+3=8

3+5=8

so,

2(8)/8

Now since there are two 8s on each side, they cancel eachother out since if you cross them over and change the sign,

2/8/8

it equals 1.

so, you are left with 2.

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Can someone please help me?
kvasek [131]

now, let's recall the rational root test, check your textbook on it.

so p  = 18 and q = 1

so all possible roots will come from the factors of ±p/q

now, to make it a bit short, the factors are loosely, ±3, ±2, ±9, ±1, ±6.

recall that, a root will give us a remainder of 0.

let us use +3.


\bf x^4-7x^3+13x^2+3x-18 \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{r|rrrrr} 3&1&-7&13&3&18\\ &&3&-12&3&18\\ \cline{1-6} &1&-4&1&6&0 \end{array}\qquad \implies (x-3)(x^3-4x^2+x+6)


well, that one worked... now, using the rational root test, our p = 6, q = 1.

so the factors from ±p/q are ±3, ±2, ±1

let's use 3 again


\bf \begin{array}{r|rrrrr} 3&1&4&1&6\\ &&3&-3&-6\\ \cline{1-5} &1&-1&-2&0 \end{array}\qquad \implies (x-3)(x-3)(x^2-x-2)


and of course, we can factor x²-x-2 to (x-2)(x+1).

(x-3)(x-3)(x-2)(x+1).

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Step-by-step explanation:

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Cloud [144]


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WILL GIVE BRAINLIEST IF CORRECT
SCORPION-xisa [38]

9514 1404 393

Answer:

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Step-by-step explanation:

Translation does not change the dimensions of the figure. The two figures together tell you the right triangle has a base of 154 units and a height of 72 units.

The area of a triangle is given by ...

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