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ratelena [41]
3 years ago
15

Which of the following correctly

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0
I think it A but I could be wrong
You might be interested in
How do u do this stuff I’m confused
Luba_88 [7]
Ok so I'll show u how to do number 7
So 12 and one third basically equals 12+(1/3)
Now in order to add this to another fraction, we need to make the mixed numbers in improper fractions
So 12 and (1/3)=(36/3)+(1/3)
And 9 and (2/3)=(27/3)+(2/3)
Ok so the first number (or 12 and (1/3)) basically equals (37/3)
The second number (or 9 and (2/3)) equals (29/3)

Now that we have the two numbers with common denominators, we can do the math.

(37/3)-(29/3) is the same as (37-29)/3
So simplifying would be
(8/3) or 2 and (2/3)
Hoped this helped
4 0
3 years ago
Given the formula x,= 2a(b-+7), find x if a=3 and- b:S
dusya [7]
2x3(S-+7)=6xS=7............
6 0
4 years ago
Evaluate using P E M D A S 2 x 3 + 4
Juli2301 [7.4K]

Step-by-step explanation:

We would do multiplication first, because of order of operations.

2x3=6

Now we add.

6+4=10

hope it helps! :3

8 0
3 years ago
Read 2 more answers
I’m not sure how to do this
seraphim [82]
\bf \cfrac{\frac{x+4}{3}+\frac{1}{x}}{5+\frac{15}{x}}\qquad \cfrac{\impliedby \frac{LCD}{3x}}{\impliedby \stackrel{LCD}{x}}\implies \cfrac{\quad \frac{x(x+4)+3(1)}{3x}\quad }{\frac{x(5)+(1)15}{x}}\implies \cfrac{\quad \frac{x^2+4x+3}{3x}\quad }{\frac{5x+15}{x}}

\bf \cfrac{x^2+4x+3}{3\underline{x}}\cdot \cfrac{\underline{x}}{5x+15}\implies \cfrac{x^2+4x+3}{3(5x+15)}\implies \cfrac{x^2+4x+3}{15x+45}
\\\\\\
\cfrac{(x+3)(x+1)}{15x+45}\implies \cfrac{\underline{(x+3)}(x+1)}{15\underline{(x+3)}}\implies \cfrac{x+1}{15}
7 0
3 years ago
Graph the line y=-3x-2
sashaice [31]

Answer:

The graph of the line is attached below.

Step-by-step explanation:

According to the slope-intercept form

y=mx+b

As the line equation  is given by

y=\:-3x\:-\:2\\

\mathrm{Slope\:of\:}-3x-2:\quad m=-3

y-intercept = b = -2

\mathrm{Axis\:interception\:points\:of}\:-3x-2:\quad \mathrm{X\:Intercepts}:\:\left(-\frac{2}{3},\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:-2\right)

  • \mathrm{x-intercept\:is\:a\:point\:on\:the\:graph\:where\:}y=0

-3x-2=0:\quad x=-\frac{2}{3}

\mathrm{X\:Intercepts}:\:\left(-\frac{2}{3},\:0\right)

  • y\mathrm{-intercept\:is\:the\:point\:on\:the\:graph\:where\:}x=0

y=-3\cdot \:0-2:\quad y=-2\\

\mathrm{Y\:Intercepts}:\:\left(0,\:-2\right)

The graph of the line is attached below.

7 0
3 years ago
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