Ask question

Ask question

All categories

2 years ago

6

The coordinates of three vertices of rectangle are A(1,-5),B(1,3),and C(10,3). Find the coordinates of the fourth vertex. Then,

find the area of the rectangle.

1 answer:

densk [106]2 years ago

6 0

The fourth coordinate is (10,-5) this is probably not correct but I don't have a graph with me.

You might be interested in

If the height is 11 cm and the area is 260.15 cm2 what is the base help please I am just in 6 grade

kogti [31]

23.65 i think beacuse i just did it in my mind unless i am totally wrong

5 0

3 years ago

Read 2 more answers

Which expression has a value greater than 253?

zzz [600]

Answer:

All of the above

Step-by-step explanation:

6 0

2 years ago

Explain how you could estimate 22% of 78 to check if your answer is reasonable.

choli [55]

you can find 10 percent by dividing 78 by 10 which is 7.8. Then you double that to find 20 percent. That will make you find 20 percent! close to 22

5 0

3 years ago

What is the area ?!!

Tresset [83]

Answer:

309.02

Step-by-step explanation:

A=2(1+2)a2=2·(1+2)·82≈309.01934

6 0

2 years ago

Read 2 more answers

Find the limit, if it exists (picture below)

valentinak56 [21]

Answer:

c. $\frac{1}{2 \sqrt{7} }$

Step-by-step explanation:

When plugging in zero into the given equation:

$\lim_{x \rightarrow 0} \frac{\sqrt{x + 7} - \sqrt{7} }{x} = \frac{0}{0}$

Answer is in indeterminate form = use L'Hospital's Rule:

(Derivative of the top / Derivative of the bottom)

$\lim_{x \rightarrow 0} \frac{ \frac{1}{2} (x + 7)^{ \frac{-1}{2}} - 0 }{1}$

Rearranged equation:

$\lim_{x \rightarrow 0} \frac{1}{2 \sqrt{x + 7} }$

Plug zero back into equation:

$\lim_{x \rightarrow 0} \frac{1}{2 \sqrt{x + 7} } = \frac{1}{2 \sqrt{0 + 7} } = \frac{1}{2 \sqrt{7} }$

Answer:

$\lim_{x \rightarrow 0} \frac{\sqrt{x + 7} - \sqrt{7} }{x} = \frac{1}{2 \sqrt{7} }$

5 0

2 years ago