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Alex Ar [27]
2 years ago
12

What is the greatest common factor of 20 and 10?

Mathematics
2 answers:
Nana76 [90]2 years ago
7 0
10 is the gcf for 20 and 10
Naddika [18.5K]2 years ago
6 0

Answer:

Step-by-step explanation:

The factors common to 20 and 10 are 1, 5 and 10.  The largest of these is 10.  That's the GCF of 20 and 10.

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The length of a rectangle is four times its width. If the perimeter is at most 130 centimeters, what is the greatest possible va
pashok25 [27]

Solution.

L = 4w --------<span>The length of a rectangle is four times its width.
</span>2w + 2L ≤ 130 ------- <span>the perimeter is at most 130 centimeters.
</span>Now, if substitute the first equation into the second inequality you will get
2w + 2 • (4w) ≤ 130.

Therefore, the inequality model in C is correct.

Bonus.
If you solve the inequality you will have a final answer w ≤ 13. The greatest value being 13.

Hope it helps,
5 0
3 years ago
Write in slope-intercept form an equation of the line that passes through the points (−6,−3) and (18,1).
Brilliant_brown [7]

6x-y+33=0

Step-by-step explanation:

m=y2-y1/x1-x2

m=1-(-3)/18-(-6)

m=4/24m=6

m=y-y1/x-x1

6=y-y1/x-x1

6x-y+33=0

4 0
2 years ago
Please Hurry!!!
Valentin [98]

The point-slope form:

y-y_1=m(x-x_1)

The formula of a slope:

m=\dfrac{y_2-y_1}{x_2-x_1}

We have the points A(-2, 6) and B(2, -2). Substitute:

m=\dfrac{-2-6}{2-(-2)}=\dfrac{-8}{4}=-2\\\\y-6=-2(x-(-2))\\\\\boxed{y-6=-2(x+2)}\to\boxed{(B)}

5 0
3 years ago
When you convert from a large unit such as pounds or minutes to a smaller unit as ounces or seconds why do you use multiplicatio
Ronch [10]
Hope this helps:))))))

6 0
3 years ago
I want help its urgent!!! <br><br>(iv), (v), (vi) Questions​
Artist 52 [7]

Solution:

iv) Given equation: x^{2} - y^{2} - 4x - 2y + 3

Since we are given two variables, we need to split the original expression into two quadratic expressions as

x^{2} - 4x + 4 - y^{2} - 2y - 1

Factoring x first:

x^{2} - 4x + 4 = x^{2} - 2x - 2x + 4

= x (x - 2) - 2 (x-2)

= (x-2)(x-2)

= (x-2)^{2}

Factoring y now:

-y^{2} - 2y - 1 = -1(y^{2} + 2y + 1)

= -1(y^{2} + y + y + 1)

= -1[ y(y+1) + 1(y+1)]

= -1 [(y+1)(y+1)]

= (y+1)(-y-1)

Therefore, the original expression becomes

(x-2)^{2}+(y+1)(-y-1)\\or \\(x-2)(x-2) + (y+1)(-y-1)

7 0
3 years ago
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