Ask question

Ask question

All categories

Keith_Richards [23]

3 years ago

9

The ordered pairs shown below follow a pattern

2 answers:

Lubov Fominskaja [6]3 years ago

5 0

Answer:

I don't see the picture I can't help p u

Amiraneli [1.4K]3 years ago

3 0

Yea there is no picture lol

You might be interested in

ABCD is a parallelogram; find the area of the shaded region.

Korolek [52]

Answer:

B. 60 ft²

Step-by-step explanation:

The overall area of the parallelogram is the product of its base length (15 ft) and height (8 ft), so is ...

overall area = (15 ft)(8 ft) = 120 ft²

The area of the quadrilateral in the middle of the parallelogram is the area of two triangles, each with a base equal to 15 ft (the horizontal line in the diagram), and heights that total 8 ft. If we let x represent the height of one of those triangles, then the other has a height of (8-x) and the sum of their areas is ...

quadrilateral area = (1/2)(15 ft)(x) +(1/2)(15 ft)(8 ft -x)

= (1/2)(15 ft)(x + 8 ft - x) = (1/2)(15 ft)(8 ft) = 60 ft²

Then the shaded area is the difference between the parallelogram area and the quadrilateral area:

shaded area = parallelogram area - quadrilateral area

= (120 ft²) - (60 ft²)

= 60 ft²

The area of the shaded region is 60 ft².

8 0

3 years ago

Rationalise the denominator of (12)/(\sqrt(10)+\sqrt(7)+\sqrt(3))

jasenka [17]

Answer:

$\frac{12}{\sqrt{10} +\sqrt{7} +\sqrt{3} }=\frac{6\sqrt{147} +6\sqrt{63}-6\sqrt{210} }{21 }$

Step-by-step explanation:

$\frac{12}{\sqrt{10} +\sqrt{7} +\sqrt{3} }$

$=\frac{12\left( \sqrt{10} -\left( \sqrt{7} +\sqrt{3} \right) \right) }{(\sqrt{10}+(\sqrt{7} +\sqrt{3 } ))( \sqrt{10} -( \sqrt{7} +\sqrt{3}))}$

$=\frac{12\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right) }{\sqrt{10^2} -(\sqrt{7} +\sqrt{3})^2}$

$\frac{12\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right) }{10-(10+2\sqrt{21} )} }$

$=\frac{12\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right) }{-2\sqrt{21} }$

$=\frac{-6\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right) }{\sqrt{21} }$

$=\frac{-6\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right) }{\sqrt{21} } \times\frac{\sqrt{21} }{\sqrt{21} }$

$=\frac{-6\sqrt{21} \left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right) }{21 }$

$=\frac{-6\sqrt{210} +6\sqrt{147} +6\sqrt{63} }{21 }$

$=\frac{6\sqrt{147} +6\sqrt{63}-6\sqrt{210} }{21 }$

<u><em>Remark</em></u>:

You can simplify moreover if you want to.

5 0

3 years ago

The area of a rectangle is 63 yd2 , and the length of the rectangle is 5 yd more than twice the width. Find the dimensions of th

Grace [21]

I hope this helps you

8 0

4 years ago

A car uses 7litres of pétrole to coker 28km . How many litres of pétrole does hé use to coker 64km

lions [1.4K]

28/7=4km is car journey covered by car in 1l

64/4=16l

5 0

3 years ago

Why can you eliminate the solution of -0.2 in the context of this problem? Check all that apply.

ivolga24 [154]

Answer: B (It does not make sense for time to be negative.) & C (The ball cannot hit the ground before it is thrown.)

Step-by-step explanation:

I answered the question already

7 0

3 years ago