3 years ago

Please help me on this

Mathematics

1 answer:

Stolb23 [73]3 years ago

5 0

Answer:

AX = 14

Step-by-step explanation:

The medians of a triangle intersect at its centroid.

The position of the centroid X from the vertex A is

AX = $\frac{2}{3}$ AL = $\frac{2}{3}$ × 21 = 14

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Solve 2x2 + 5x = 12

aliya0001 [1]

2x2+5x=12, 2x+5x=7x, 12+2=14, 7x/7=14/7, x=2

4 0

4 years ago

X + 2 > 7 or 4 - x > 6<br><img src="https://tex.z-dn.net/?f=x%20%2B%202%20%3E%207or4%20-%20x%20%3E%206%20" id="TexFormula1

Doss [256]

So, x > 7 - 2; x >5 or

4 - x > 6; 4 - x - 6 > 0;

4 - 6 -x > 0;

- 2 -x > 0;

-2 > x;

Finally, x > 5 or x < -2;

4 0

3 years ago

on monday, 251 students went on a trip to the zoo.all 5 buses were filled and 6 students had nto travel in cars.how many student

Norma-Jean [14]

Answer: 49

Step-by-step explanation:

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5 0

3 years ago

The number of cars at a dealership dropped from 64 to 48 after a weekend sale.

Arada [10]

The answer is 25%

(1)
If 64 is 100%, 48 is how much percent:
64 : 100% = 48 : x
x = 100% * 48 : 64
x = 75

(2)
Percent of decrease is:
100% - x = 100% - 75% = 25%

7 0

3 years ago

What is x, the distance between points A and A'? 2. 4 units 4. 8 units 13. 6 units 14. 4 units.

adelina 88 [10]

The distance between points A (3, -5) and A' (2, -3) is 2.4 units.

Given that,

The points A (3, -5) and A' (2, -3).

We have to determine,

The distance between point A and A'.

According to the question,

The distance between two points is determined by using the distance formula.

$\rm Distance \ formula = \sqrt{(x_2-x_2) ^2+ (y_2-y_2)^2$

Then,

The distance between points A (3, -5) and A' (2, -3) is,

$\rm Distance = \sqrt{(2-3) ^2+ ((-3)-(-5))^2}\\\\ Distance = \sqrt{(-1) ^2+ (2)^2}\\\\ Distance = \sqrt{1+4}\\\\Distance = \sqrt{5}\\\\Distance = 2.4 \ units$

Hence, The distance between points A (3, -5) and A' (2, -3) is 2.4 units.

For more details refer to the link given below.

brainly.com/question/8069952

6 0

2 years ago

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