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Lina20 [59]
3 years ago
13

Using the distributive property, break apart 6x4 = 24

Mathematics
1 answer:
____ [38]3 years ago
8 0
6x8=
{6x 4} + {6x 4 }
= + =
24 24
48
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SOMEONE HELP ME PLZ ASAP!!!!​
Elanso [62]

A=22cm

B=21cm

C=23cm

Good luck

4 0
3 years ago
Explain why a set of points that defines a plane cannot be collinear
USPshnik [31]

Answer:

Three non-collinear points determine a plane.

Step-by-step explanation:

This statement means that if you have three points not on one line, then only one specific plane can go through those points. The plane is determined by the three points because the points show you exactly where the plane is

3 0
3 years ago
Find the following rates. Round your answer to the nearest hundredth. a. ? % of 75 = 5 b. ? % of 28 = 140 c. ? % of 100 = 40 d.
Nimfa-mama [501]
A.15 b.500 c.40 d.7.5 yeah, not to confident but I think it's correct.
5 0
3 years ago
4/x-1+6/x+1=-12/x^2-1
Mademuasel [1]
\frac{4}{x - 1} + \frac{6}{x + 1} = \frac{-12}{x^{2} - 1}

\frac{4(x + 1)}{(x - 1)(x + 1)} + \frac{6(x - 1)}{(x - 1)(x + 1)} = \frac{-12}{x^{2} - 1}

\frac{4x + 4}{x^{2} - 1} + \frac{6x - 6}{x^{2} - 1} = \frac{-12}{x^{2} - 1}

\frac{10x - 2}{x^{2} - 1} = \frac{-12}{x^{2} - 1}

(10x - 2)(x^{2} - 1) = -12(x^{2} - 1)
10x^{3} - 10x^{2} - 2x^{2} + 2 = -12(x^{2}) + 12(1)
10x^{3} - 12x^{2} + 2 = -12x^{2} + 12
10x^{3} - 12x^{2} + 12x^{2} + 2 = -12x^{2} + 12x^{2} + 12
10x^{3} + 2 - 2 = 12 - 2
\frac{10x^{3}}{10} = \frac{10}{10}

x^{3} = 1
x = 1
4 0
3 years ago
Is the line through points P(0, –9) and Q(2, –8) perpendicular to the line through points R(1, 4) and S(3, 3)? Explain
Serjik [45]
The quickest way to tell if a line is perpendicular is to find the slope, so you'll need to get the slope for both lines

PQ = ((-8) - (-9))/(2 - 0) = 1/2

and, for reference, a perpendicular slope is the negative reciprocal of the original (so we're looking for -2)

RS = (3 - 4)/(3 - 1) = -1/2

so, no, these two lines aren't perpendicular because the slope of RS is only the negative of PQ, not the negative <em>reciprocal</em>.
3 0
3 years ago
Read 2 more answers
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