Figure is not d

Mathematics

1 answer:

ollegr [7]2 years ago

7 0

Answer:

1.5

Step-by-step explanation:

the feet in perimeter is going to be the same

You might be interested in

Hello, can you explain this topic im in 6th grade

Andrew [12]

Hello there I hope you are having a great day :)

The Topic Shifting digits:

Shifting digits - Putting a digit in a different place. For a example Take the number 358 you would exchange the number from the hundreds that would equal 838 this meaning that you what to make a bigger number than a small one.

Examples:

1. 358 equal 838

2. 204 equal 402

3. 188 equal 881

You are just making them smaller to a greater number :)

Hopefully that helps you :)

5 0

2 years ago

Read 2 more answers

49.50 increased with 55% discount of 51%=

saw5 [17]

the answer is $37.60

5 0

2 years ago

Rachel runs 3.2 miles each weekday and 1.5 miles each day of the weekend how many miles will she have run in six weeks

elena-14-01-66 [18.8K]

3.2 x 5 + 1.5 x 2 = 19

4 0

2 years ago

87+25 <img src="https://tex.z-dn.net/?f=12%20%2B%2010%20%20" id="TexFormula1" title="12 + 10 " alt="12 + 10 " align="absmid

Butoxors [25]

Answer:

Seriously?

U don't know the answer

3 0

2 years ago

Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3

omeli [17]

$\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}$

$=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1$

$=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

3 0

2 years ago