What is the GCF of 2x^6-12x^4

Olivia wants to walk 8000 steps today. She walked 45% of this total before lunch? How many steps did Olivia walk before lunch?

LiRa [457]

Answer:

3600 steps

Step-by-step explanation:

Step 1: Multiply 45 by 8000.

$45 * 8000$
$360000$

Step 2: Move the decimal point two places to the left.

$36000.0$
$3600.00$

Therefore, the answer is 3600 steps.

Have a lovely rest of your day/night, and good luck with your assignments! ♡

4 0

2 years ago

ILL MARK BRAINLEST When Josiah opens and lays a cereal box out flat, he sees that the top and the bottom of the box both measure

Elan Coil [88]

Answer:

Surface area = 252 inches^2

Step-by-step explanation:

Surface area = 2lw + 2lh + 2wh

2(2*8) + 2(2*11) + 2(8*11)

32 + 44 + 176 = 252

4 0

3 years ago

lauren wants to keep her cell phone bull under $60 per month. Her current cell phone plan is $30 per month plus $0.50 per text.

stellarik [79]

Answer:

The correct answer is t < 60.

Step-by-step explanation:

Lauren wants to keep her cell phone bill below $60 per month.

Lauren's current cellphone plan charges her a fixed price of $30 and per text price for one text is $0.50.

Let Lauren sends t texts in a complete month.

Total money spent on texts in a month is given by $ (0.50 × t)

Therefore Lauren's total spent in a month is given by $ (30 + (0.50 × t)).

But this amount should be under $60 as per as the given problem.

∴ 30 + (0.50 × t) < 60

⇒ (0.50 × t) < 30

⇒ t < $\frac{30}{0.50}$

⇒ t < 60.

So in order to keep her phone monthly bill under $60, Lauren should keep her number of texts below 60.

8 0

2 years ago

There are 3900 workers in the three main buildings downtown. Twice as many people work in the largest building as in the smalles

Aleks04 [339]

There are 850 workers in the smallest building, 1350 workers in the second-largest building, and 1700 workers in the largest building.

7 0

3 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