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Tema [17]
3 years ago
5

Which answer shows this equation written in standard form?

Mathematics
2 answers:
STatiana [176]3 years ago
5 0
4-3(x-y)=-8x+2
=4-3x-3y=-8x+2
=4-3x+8x-3y-2
=5x-3y+2
or 
5x-3y=-2
so it will be C=5x+3y=-2
Elden [556K]3 years ago
4 0
To write the expression to its standard form, we have to simplify the expression given. The standard form of the given should be the standard equation for the equation of a line which is:

ax + by = c

<span>4 – 3(x – y) = –8x + 2
</span>4 - 3x + 3y = -8x +2
- 3x + 3y +<span>8x</span> = 2 - 4
5x + 3y = -2

Therefore, the correct answer among the choices is option C.
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Walt received a package that is 2 1/3 inches long, 6 3/4 inches high, and 8 1/2 inches wide. What is the surface area of the pac
grin007 [14]
Add whole numbers: 2 + 6 + 8 = 16
Add fractions:

= 1/3 + 3/4 + 1/2

= (8 + 18 + 12) ÷ 24

= 38/24 or 1 14/24 or simplified to 1 7/12

Total surface area = 16 + 1 7/12 or 17 7/12


5 0
3 years ago
Read 2 more answers
HELP PLEASE I NEED TO TURN THIS IN BEFORE WEDNESDAY
JulijaS [17]

Answer:

451

Step-by-step explanation:

4 0
2 years ago
Please give me the correct answer.​
nydimaria [60]

Answer:

The third one which is the one on the bottom to the right under 34 the ont he bottom

7 0
3 years ago
CAN SOMEONE PLEASE HELP ME WORTH 20 points
Ainat [17]

Answer:

  12 units

Step-by-step explanation:

Point H is the intersection of the medians, so divides each median into parts with the ratio 2:1. Here, that is ...

  CH : HG = 2 : 1 = CH : 6

Then CH = 2×6 = 12.

_____

Alternatively, you could determine altitude CG from the Pythagorean theorem applied to triangle BCG. You would compute ...

  CG² +GB² = BC²

  CG = √(BC² -GB²) = √(19.8² -8²) = √146.7 ≈ 12.1

This is an approximation based on the fact that the value 9.9 for BF is an approximation. Based on BG and GH being integer values, the appropriate measure of BF is (√388)/2 ≈ 9.8489.

(The triangle shown cannot exist.)

5 0
3 years ago
Simplify
OlgaM077 [116]

Answer:

see below

Step-by-step explanation:

\frac{x-1}{x^2-3x+2}+ \frac{x-2}{x^2-5x+6} +\frac{x-5}{x^2-8x+15}

we need to simplify that

x^2-3x+2=(x-1)(x-2)\\\\x^2-5x+6=(x-2)(x-3)\\\\x^2-8x+15=(x-3)(x-5)

so we can continue

\frac{x-1}{(x-1)(x-2)}=\frac{1}{x-2}\\\\\frac{x-2}{(x-2)(x-3)} =\frac{1}{x-3}\\\\\frac{x-5}{(x-3)(x-5)} =\frac{1}{x-3}

and we can put all together

\frac{1}{x-2}+ \frac{1}{x-3}+ \frac{1}{x-3}\\\\\frac{1}{x-2} +\frac{2}{x-3}\\\\\frac{x-3}{(x-3)(x-2)}+ \frac{2(x-2)}{(x-2)(x-3)} \\\\\frac{x-3+2x-4}{(x-3)(x-2)}\\\\\frac{3x-7}{x^2-5x+6}

3 0
3 years ago
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