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KonstantinChe [14]
2 years ago
10

2 and 5. What is the LCM?

Mathematics
1 answer:
Masja [62]2 years ago
5 0
Well, first of all, LCM means Lowest Common Multiple. So you have to write out the factors of each of the numbers. So that will be;
2= 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
5=5, 10, 15, 20
  So now, you have to pick out the numbers that are common to both numbers. So;
10, 20
  10 and 20 are the only common numbers. So you have to look for the lowest common number. And that is 10. So the LCM is 10.. Hope i helped. Have a nice day. 
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PLEASE ANSWER !!
ANEK [815]

Answer:

Check the explanation.

Step-by-step explanation:

As the graph of a linear function f passes through the point (-2,-10) and has a slope of 5/2.

As the slop-intercept form is given by:

y = mx+b

where m is the slope and b is the y-intercept.

substituting the values (-2, -10) and m = 5/2 in the slop-intercept form to determine y-intercept.

y = mx+b

-10=\frac{5}{2}\left(-2\right)+b

-\frac{5}{2}\cdot \:2+b=-10

-5+b=-10

-5+b+5=-10+5

b=-5

And the equation of the line in the slope-intercept form will be:

y = mx+b

putting b = -5 and slope = m = 5/2

y = mx+b

y=\frac{5}{2}x+\left(-5\right)

y=\frac{5}{2}x-5

Determining the zero of function.

As we know that the real zero of a function is the x‐intercept(s) of the graph of the function.

so let us determine the value of x (zero of function) by setting y = 0.

0=\frac{5}{2}x-5

\frac{5}{2}x-5=0

\frac{5}{2}x=5

5x=10

x=2

Therefore, the zeros of the function will be:

  • x = 2
3 0
3 years ago
What time is 5 3/4 hours after 9:22pm
12345 [234]
2:40 I think, but I may be wrong
5 0
2 years ago
Read 2 more answers
Q3.
Bas_tet [7]
1-560 WOULD BE THW ANSWER IM SORRY THIS IS
8 0
2 years ago
Which number one shows Point A at-2, Point B at 35, Point Cat-3
kap26 [50]

Answer:

B

D

A

с

2

2-1

1

2

3 4 5

2 1

2

2 4 5

A

B

2-1

1

2

B

5 4 3 2 1

1

2 3 4 5

5 0
2 years ago
Webassign find the area of the region bounded by the parabola y = x2, the tangent line to this parabola at the point (6, 36), an
Dvinal [7]
First find the tangent line

dy/dx=2x
at x=6, the slope is 2(6)=12
so
use point slope form
y-y1=m(x-x1)
point is (6,36)
so
y-36=12(x-6)
y-36=12x-72
y=12x-36

alright, so we know they intersect at x=6
and y=12x-36 is below y=x^2

so we do \int\limits^6_0 {x^2} \, dx - \int\limits^6_0 {12x-36} \, dx = \int\limits^6_0 {x^2-(12x-36)} \, dx = \int\limits^6_0 {x^2-12x+36} \, dx =
[\frac{x^3}{3}-6x^2+36x]\limits^6_0=(\frac{6^3}{3}-6(6)^2+36(6))-(0)=\frac{216}{3}-216+216= 72+0=72

the area under the curve bounded by the lines and the x axis is 72 square units
4 0
2 years ago
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