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3 years ago

Where could point E be and D is between E and B?

Mathematics

1 answer:

Butoxors [25]3 years ago

6 0

It is d.2. because its like a pattern you have to follow on the graph.

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What is the area of the shape?

marshall27 [118]

Step-by-step explanation:

hello,

the shape is a right trapezium.

you use the following formula:

area of a trapezium = ½ × sum of parallel sides× height.

or simply,

a = ½ ( a + b ) h

where a and b are the parallel sides

and h is the height.

thus,

a = ½ × ( 4 + 6 )× 4

a = ½ ( 10 ) × 4

a = 5 × 4

a = 20

have a great day! au revoir !

6 0

2 years ago

Read 2 more answers

4√6 •√3 how do I show work for this because the answer is 2√12

Yuliya22 [10]

Answer:

$4 \sqrt{6} \times \sqrt{3} = 12 \sqrt{2}$

Step-by-step explanation:

We want to simplify the radical expression:

$4 \sqrt{6} \times \sqrt{3}$

We write √6 as √(2*3).

This implies that:

$4 \sqrt{6} \times \sqrt{3} = 4 \sqrt{2 \times 3} \times \sqrt{3}$

We now split the radical for √(2*3) to get:

$4 \sqrt{6} \times \sqrt{3} = 4 \sqrt{2} \times \sqrt{3} \times \sqrt{3}$

We obtain a perfect square at the far right.

$4 \sqrt{6} \times \sqrt{3} = 4 \sqrt{2} \times (\sqrt{3} )^{2}$

This simplifies to

$4 \sqrt{6} \times \sqrt{3} = 4 \sqrt{2} \times 3$

This gives us:

$4 \sqrt{6} \times \sqrt{3} = 4 \times 3 \sqrt{2}$

and finally, we have:

$4 \sqrt{6} \times \sqrt{3} = 12 \sqrt{2}$

5 0

3 years ago

Please help me!!! almost done but can’t find this answer!!! thanks

ivolga24 [154]

A, C & E!!! your welcome btw

4 0

3 years ago

½ (6x + 8) = 2x + 4 + x

bearhunter [10]

Answer:

True

Step-by-step explanation:

Both sides (right side and left side) of the equation are indeed equal.

I hope this helps!

5 0

3 years ago

Read 2 more answers

The domain of y=ax^2 is all real numbers. Describe the range when (a) a>0 and (b) a<0.

eduard

The range is all positive values...

R{y | y>0}

Range > "Y" such that y is greater than 0

3 0

3 years ago