All categories

3 years ago

What is the common factor of 49 and 50?

Mathematics

2 answers:

Zinaida [17]3 years ago

6 0

The factors of 49 are: 1, 7 and 49.
The factors of 50 are: 1, 2, 5, 10, 25 and 50.

The common factor is 1.

Reika [66]3 years ago

6 0

$49|7\\ 7 \ |7 \\ 1 \\ \\ 49=7^2 \\ \\ 50|2\\25|5\\5\ |5 \\1\\\\50=2*5^2 \\ \\ G.C.F(49;50)=2450$

You might be interested in

Select the values that make the inequality 4q≥−68 true. Then write an equivalent inequality, in terms of q. (Numbers written in

Rina8888 [55]

Answer:

-27 = q

Step-by-step explanation:

I am not sure that my answer is correct but hopefully that helps.

8 0

3 years ago

The distance between Maggie’s home and her school is 5,280 yards. (Hint: 1 mile = 1,760 yards.)

Mariulka [41]

1 mile = 1760 yards
5280 yards = 5280 / 1760 = 3 miles

4 0

3 years ago

1. En una tienda de ropa se pusieron en oferta guantes y chamarras. El primer dia se vendieron 6 pares

Softa [21]

10560 so much money all forget to count

6 0

2 years ago

Read 2 more answers

Solve for g.

Y_Kistochka [10]

Answer:

g =6/5 g=-5/2

Step-by-step explanation:

(5g − 6)(2g + 5) = 0

Using the zero product property

(5g − 6) =0 (2g + 5) = 0

5g-6+6 =0+6 2g+5-5=0-5

5g =6 2g =-5

5g/5 = 6/5 2g/2 =-5/2

g =6/5 g=-5/2

7 0

3 years ago

Find+the+positive+value+for+α+if+the+radius+of+the+circle+3x^2+3y-6αx+12y-3α=0+is+4

Alik [6]

Answer:

$\alpha=3$

Step-by-step explanation:

<u>Equation of a Circle</u>

A circle of radius r and centered on the point (h,k) can be expressed by the equation

$(x-h)^2+(y-k)^2=r^2$

We are given the equation of a circle as

$3x^2+3y^2-6\alpha x+12y-3\alpha=0$

Note we have corrected it by adding the square to the y. Simplify by 3

$x^2+y^2-2\alpha x+4y-\alpha=0$

Complete squares and rearrange:

$x^2-2\alpha x+y^2+4y=\alpha$

$x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4$

$(x-\alpha)^2+(y+2)^2=r^2$

We can see that, if r=4, then

$\alpha+\alpha^2+4=16$

Or, equivalently

$\alpha^2+\alpha-12=0$

There are two solutions for $\alpha$ :

$\alpha=-4,\ \alpha=3$

Keeping the positive solution, as required:

$\boxed{\alpha=3}$

8 0

3 years ago