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2 years ago

2 and 5. What is the LCM?

1 answer:

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:

2-1

3 4 5

2 1

2 4 5

2-1

5 4 3 2 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