2 years ago

What is the area of the triangle whose vertical are (-4,0), (-10,0) and (-7,-6)?

Mathematics

1 answer:

olchik [2.2K]2 years ago

8 0

Step-by-step explanation:

from the point -10 to -4 thre is a difference of 6 uints . Base 6

from the x-axis to furthest point,

there is a difference of 6 units. Height 6

You might be interested in

Hannah can complete 4 math problems per minute.write and solve a direct variation equation to find how many math problems she ca

scoundrel [369]

15 x 4 = 60
Hope this helps!

4 0

4 years ago

Find the 81st term of the arithmetic sequence -10, -25, -40

Maurinko [17]

Answer:

5 - (N x 15)

81 x 15 = 1215

5-1215=-1210

-1210 should be the 81st term

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The answer is red show step by step of the pro tem to get the answer

miskamm [114]

In order to rationalize the denominator of each expression, we need to multiply the expression by the same radical in the denominator, this way we can remove the radical from the denominator.

$\frac{5\sqrt{4}}{\sqrt{3}}(\cdot\sqrt{3})=\frac{5\sqrt{4}\sqrt{3}}{(\sqrt{3})^2}=\frac{5\cdot2\cdot\sqrt{3}}{3}=\frac{10\sqrt{3}}{3}$

10)

$-\frac{5}{3\sqrt{2}}(\cdot\sqrt{2})=-\frac{5\sqrt{2}}{3(\sqrt{2})^2}=-\frac{5\sqrt{2}}{3\cdot2}=-\frac{5\sqrt{2}}{6}$

11)

$\frac{2\sqrt{3}}{4\sqrt{5}}(\cdot\sqrt{5})=\frac{2\sqrt{3}\sqrt{5}}{4(\sqrt{5})^2}=\frac{2\sqrt{15}}{4\cdot5}=\frac{\sqrt{15}}{2\cdot5}=\frac{\sqrt{15}}{10}$

12)