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3 years ago

Given the range {-1, 5, 9), what is the domain for the relation y = x + 3?

Mathematics

2 answers:

denis23 [38]3 years ago

8 0

Answer:

yes

Step-by-step explanation:

because it happens that way that it wants

andrezito [222]3 years ago

5 0

Answer:

The answer is {-4,2,6}

Step-by-step explanation:

I am sorry if i had more time i would explain but i gotta go.

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I need help with math

pantera1 [17]

Answer:

answer is y=-2x

Step-by-step explanation:

-2x because every time the dot decreases

6 0

2 years ago

Use the figure below. Which angle is congruent to ∠ACB? A. ∠ABC B. ∠AED C. ∠ADE D. ∠DAE

Basile [38]

Answer:

Angle ABD

Step-by-step explanation:

I think it is so because the angles are aligned already so you what has the same value.

5 0

2 years ago

Read 2 more answers

What is 6,4 rounded to the nearest whole number

Aleks04 [339]

6.4 rounded to the nearest whole number will be 6.

if the decimal was 5 or larger, it would’ve been 7.

6 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

3 years ago

Plz help <br>2×3+6×9×2×1×4-68-1+86=<br><br>plz help

nordsb [41]

The final answer would be 455

5 0

2 years ago