All categories

3 years ago

Common factor of 6 Abd 12

Mathematics

2 answers:

Keith_Richards [23]3 years ago

5 0

The common factor is 6

nevsk [136]3 years ago

4 0

the answer to your problem is 6

You might be interested in

A translation moves A(-3,2) to a'(0,0). find b', the image of b(5,4)) under the same translation

statuscvo [17]

The image of b, b', would be (8,2)

hope this helps!

7 0

3 years ago

If you have 8/12 of a pizza, how many thirds do you have?

saw5 [17]

You would have 2/3 because 4 can go into the numerator 2 which makes the numerator of your answer 2, then 4 goes into 12 4 Times which makes the denominator of your answer which is 3 so 2/3

5 0

3 years ago

Read 2 more answers

A consumer charges a 2530.16 purchase on the credit card the card has a daily interest rate of .042% if the consumer piece of th

svp [43]

Answer:

$31.88

Step-by-step explanation:

Given

Consumer charges = 2530.16cents

Consumer charges in dollars = $25.3016

Interest rate per day = 0.042%

Interest gotten per day = 0.042×25.3016

Interest per day = $1.06266

Interest in 30 days = ,30×1.06266

Interest in 30 days = $31.88

3 0

3 years ago

What is 5.75 roy to the nearest tenth

Furkat [3]

5.8 or 5.80 it will be the same thing tho

5 0

3 years ago

Read 2 more answers

PLS HELP BRAINLEIST AND 25 POINTS

kramer

Given:

The vertices of a triangle ABC are A( 0, 2), B (2, 5), and C (−1, 7).

To find:

Part A: The length of each side of the triangle.

Part B: The slope of each side of the triangle.

Part C: Classify the triangle.

Solution:

Part A: The vertices of a triangle ABC are A( 0, 2), B (2, 5), and C (−1, 7).

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using the distance formula, we get

$AB=\sqrt{(2-0)^2+(5-2)^2}$

$AB=\sqrt{2^2+3^2}$

$AB=\sqrt{4+9}$

$AB=\sqrt{13}$

Similarly,

$BC=\sqrt{\left(-1-2\right)^2+\left(7-5\right)^2}$

$BC=\sqrt{13}$

And,

$AC=\sqrt{\left(-1-0\right)^2+\left(7-2\right)^2}$

$AC=\sqrt{26}$

Therefore, the length of each side of the triangle are $AB=\sqrt{13},BC=\sqrt{13},AC=\sqrt{26}$ .

Part B:

Slope formula:

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

Slope of side AB is:

$m_{AB}=\dfrac{5-2}{2-0}$

$m_{AB}=\dfrac{3}{2}$

Slope of side BC is:

$m_{BC}=\dfrac{7-5}{-1-2}$

$m_{BC}=\dfrac{2}{-3}$

$m_{BC}=-\dfrac{2}{3}$

Slope of side AC is:

$m_{AC}=\dfrac{7-2}{-1-0}$

$m_{AC}=\dfrac{5}{-1}$

$m_{AC}=-5$

Therefore, the slopes of sides AB, BC, AC are $\dfrac{3}{2},-\dfrac{2}{3},-5$ respectively.

Part C:

$AB=BC=\sqrt{13}$

Product of slopes of AB and BC is:

$\dfrac{3}{2}\times \dfrac{-2}{3}=-1$

So, side AB and BC are perpendicular to each other.

Two sides of triangle ABC are equal and one angle is a right angle.

Therefore, the triangle ABC is an isosceles right triangle.

4 0

3 years ago

Read 2 more answers