Guest #3877

2 months ago

Use foil, (first outside, inside, last) and simplify to get: x^2+3x+2

Question

Tyra is training for a bicycle race. Each week she rides a total distance greater than 10 kilometers and less than or equal to 30 kilometers. If the distance is always an even number and a multiple of 3, what are the possible distances Tyra rides in one week?

Solution 1

12 km,18 km,24 km,30 km only this are possible.

(i think)

(i think)

Question

there are 54 people at the party. 18 of them are wearing red, what percent of them are not wearing red

Solution 1

**There are 54 people at the party.**

**18 people are wearing red which means that 36 people are not wearing red**

**54 - 18 = 36**

The fraction represents the 36 people at the party that are wearing no red. We can find what percent of the people at the party are not wearing red by turning into a percent.

can be reduced to Â by dividing both the numerator and denominator by the greatest common factor of 36 and 54.

2 Ã· 3 = 0.667

0.667 Ã— 100 = 66.67%

Therefore, 66.67% of the people at the party are not wearing red

Solution 2

18 = number wearing red

54-18 = 36 not wearing red

36 out of 54 = 36/54

36/54 = 0.6666

0.6666 * 100 = 66.67%

Percent wearing red is 66.67%

54-18 = 36 not wearing red

36 out of 54 = 36/54

36/54 = 0.6666

0.6666 * 100 = 66.67%

Percent wearing red is 66.67%

Question

A wooden peg game in the shape of a triangular prism is 2 inches tall. The triangle has a base of 12 inches and a height of 9 inches. Find the volume of the game.

Solution 1

The volume of the game will be given by:

Volume=(base area)Ã—height

base area=1/2Ã—baseÃ—height

Â Â Â Â Â Â Â Â =1/2Ã—12Ã—9

Â Â Â Â Â Â Â Â =54 inÂ²

Therefore the volume will be:

Volume=54Ã—2=108 inÂ³

Volume=(base area)Ã—height

base area=1/2Ã—baseÃ—height

Â Â Â Â Â Â Â Â =1/2Ã—12Ã—9

Â Â Â Â Â Â Â Â =54 inÂ²

Therefore the volume will be:

Volume=54Ã—2=108 inÂ³

Question

Whatâ€™s the ratio decimal of
2:3
3:7
3:6
3:8
2:5
3:7
2:4

Solution 1

To find the ratio decimal form, write each ratio as it's corresponding fraction, then divide. Simplify fractions, if possible.

2:3= 2/3= 0.6666666= 0.67 rounded

3:7= 3/7= 0.42857= 0.43 rounded

3:6= 3/6= 1/2= 0.5

3:8= 3/8= 0.375

2:5= 2/5= 0.4

3:7= 3/7= 0.42857= 0.43 rounded

2:4= 2/4= 1/2= 0.5

Hope this helps! :)

2:3= 2/3= 0.6666666= 0.67 rounded

3:7= 3/7= 0.42857= 0.43 rounded

3:6= 3/6= 1/2= 0.5

3:8= 3/8= 0.375

2:5= 2/5= 0.4

3:7= 3/7= 0.42857= 0.43 rounded

2:4= 2/4= 1/2= 0.5

Hope this helps! :)

Question

the greatest common factor gcf of 2 monomiald is 3x^2y. One of the monomials is 3x^4y . Which couod be the other monomial.

Solution 1

While I cannot see your answer choices, the correct answer would be one that has a coefficient that is a multiple of 3 and has an x raised to an exponent of no more than 2.Â It could have any other variables in it.

The reason this has to be is because of the GCF.Â Since 3 is part of the GCF, and the other monomial has a coefficient of 3, this means the missing monomial must be divisible by 3 as well.

Since xÂ² is part of the GCF, the missing monomial must contain xÂ².Â However, if it were to contain xÂ³ or anything larger, the GCF would have x with a higher exponent.

The reason this has to be is because of the GCF.Â Since 3 is part of the GCF, and the other monomial has a coefficient of 3, this means the missing monomial must be divisible by 3 as well.

Since xÂ² is part of the GCF, the missing monomial must contain xÂ².Â However, if it were to contain xÂ³ or anything larger, the GCF would have x with a higher exponent.

Question

how do I solve a problem in which radicals with unlike terms are added? E.g. (Square root of 3) + (two times the square root of 108) + (four times the square root of 75)

Solution 1

By making all three roots to be square roots of 3. And then add them.

Question

Which graph is the solution to the system y > 2x â€“ 3 and y < 2x + 4?

Solution 1

The solution **region **of the inequalities y > 2x - 3 and y < 2x + 4 would be the region between the parallel lines y = 2x - 3 and y = 2x + 4.

Linear Inequalities are the statements in mathematics where the left and right hand side are separated using inequality **symbols **like <, >, â‰¤ and â‰¥.

We have the **system **of inequalities given here:

y > 2x - 3 and

y < 2x + 4

To solve this first take the inequalities as **equations **y = 2x - 3 and y = 2x + 4.

Take y = 2x - 3

When x = 0, then y = -3

When x = 1, then y = -1

When y = 0, then x = 1.5

We get three **points **here (0, -3), (1, -1) and (1.5, 0).

Draw a line passing through these points.

**Substitute **(x, y) as (0, 0), then the inequality y > 2x - 3 become 0 > -3, which is true. Therefore solution region is the region containing the origin.

Similarly, take y = 2x + 4.

When x = 0, then y = 4

When x = 1, then y = 6

When y = 0, then x = -2

We get three points here, (0, 4), (1, 6) and (-2, 0).

Substitute (x, y) as (0, 0), then the inequality y < 2x + 4 become 0 < 4, which is true. Therefore solution region is the region containing the **origin**.

Hence the **solution **region of these inequalities would be the region lower to the line y = 2x + 4 and the region upper to the line y = 2x - 3.

To learn more about **Linear Inequalities**, click:

#SPJ3

Solution 2

**S****e****c****o****n****d**** **option on e2020

**Step-by-step explanation:**

Trust mee

Question

What is a 4:1 ratio? Is it equivalent to 75% to 25%?

Solution 1

Ratios are basically:

for this much of this:theres this much of that

lets say you have 4:1 in spoons and forks.

for every 4 spoons, there is 1 fork. if there is 8 spoons there are 2 forks and so on.Â Â

The percent is 400%

if you meant 1:4 the percent would be 25%. all other rules apply.Â

for every 1 spoon you have 4 forks.

for this much of this:theres this much of that

lets say you have 4:1 in spoons and forks.

for every 4 spoons, there is 1 fork. if there is 8 spoons there are 2 forks and so on.Â Â

The percent is 400%

if you meant 1:4 the percent would be 25%. all other rules apply.Â

for every 1 spoon you have 4 forks.

Question

99 Points Need Help! How to reflect over y= -x+6

Solution 1

To reflect in the line y = -x + 6, we translate everything down 6 first. This will make it seem like we are reflecting in the line y = -x

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (x,y) â†’ (x, y-6)

Then, to reflect in the line y = -x, we switch the x- and y-coordinates and then make them negative

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (x,y) â†’ (-(y-6), -x)Â Â Â Â Â

Then we move everything back up again

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (x,y) â†’ (-(y-6), -x + 6)Â Â Â Â Â Â Â Â Â

I will present to you an example. Reflect the point (-4, 8) in the line y = -x + 6.

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (-4,8) â†’ (-(8-6), -(-4) + 6) â†’ (-2, 10)

You should graph this out to confirm with the reflection line.

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (x,y) â†’ (x, y-6)

Then, to reflect in the line y = -x, we switch the x- and y-coordinates and then make them negative

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (x,y) â†’ (-(y-6), -x)Â Â Â Â Â

Then we move everything back up again

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (x,y) â†’ (-(y-6), -x + 6)Â Â Â Â Â Â Â Â Â

I will present to you an example. Reflect the point (-4, 8) in the line y = -x + 6.

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (-4,8) â†’ (-(8-6), -(-4) + 6) â†’ (-2, 10)

You should graph this out to confirm with the reflection line.

Solution 2

** Â Â (x,y) â†’ (-(y-6), -x + 6) Â **

Question

Please Help!!!!!! So i know how to reflect over line y=-x but how would i do it over line y=-x+6. An example would be helpful and a rule for more problems of this kind.

Solution 1

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (x,y) â†’ (x, y-6)

Then, to reflect in the line y = -x, we switch the x- and y-coordinates and then make them negative

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (x,y) â†’ (-(y-6), -x)Â Â Â Â Â

Then we move everything back up again

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (x,y) â†’ (-(y-6), -x + 6)Â Â Â Â Â Â Â Â Â

I will present to you an example. Reflect the point (-4, 8) in the line y = -x + 6.

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (-4,8) â†’ (-(8-6), -(-4) + 6) â†’ (-2, 10)

You should graph this out to confirm with the reflection line.

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