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3 years ago

10

When two fractions are between 0 and half how do you know which fraction is greater explain

2 answers:

Sphinxa [80]3 years ago

8 0

1/4 and 1/15. 1/4 is greater because there is less fourths in a whole than fifteenths.

nevsk [136]3 years ago

6 0

To know which fraction is greater than another is to first look at the denominator

<u>example</u>
$\frac{1}{5} > \frac{1}{8}$

if u look at the denominator the one with the highest denominator is the smallest and then you look at the numerator to imagine if its closer to the denominator or farther away from the denominator.

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Jose has saved $75 to take tutoring classes. He must pay a registration fee of $15 and then an additional $6 for each class.

bonufazy [111]

Answer:

15+6x=75

Step-by-step explanation:

15 is the starting fee and it is 6$ per class so it would be 6x+15=75

4 0

3 years ago

Read 2 more answers

Ali will rent a car for the weekend. He can choose one of two plans. The first plan has an initial fee of 70 and costs an additi

Aleksandr-060686 [28]

Answer: Ali would need to drive 350 miles for the two plans to cost the same.

Step-by-step explanation:

This question can be solved by creating two equations using the information supplied in the question and then solving these simultaneously.

Let the cost be C.

Let the number of miles be M.

Let the initial payment be i.

Let the rate per mile driven be R.

Plan 1:

C = i+R×M

C = 70+0.60M ... equation 1

Plan 2:

C = i+R×M

C = 0+0.80M

C = 0.80M ...equation 2

Substituting equation 2 into equation 1:

0.80M = 70+0.60M

0.80-0.60M = 70

0.20M = 70

M = 70/0.20

M = 350 miles

5 0

3 years ago

Point T is at (-3, 8). What are the coordinates of T' after R(y-axis) o R(x-axis)?

Whitepunk [10]

Given:

Point is T(-3,8).

To find:

The coordinates of T' after $R(y-axis)\circ R(x-axis)$ .

Solution:

We know that, $R(y-axis)\circ R(x-axis)$ means the figure reflected across the x-axis then reflected across y-axis.

If a figure reflected across x-axis, then

$(x,y)\to (x,-y)$

$T(-3,8)\to T_1(-3,-8)$

If a figure reflected across y-axis, then

$(x,y)\to (-x,y)$

$T_1(-3,-8)\to T'(-(-3),-8)$

$T_1(-3,-8)\to T'(3,-8)$

Therefore, the required point is T'(3,-8).

4 0

3 years ago

Can anyone solve this? will give brainliest if correct.

miv72 [106K]

Answer:

See below

Step-by-step explanation:

<u>The missing reasons are:</u>

1. k. Given
2. j. Definition of parallelogram
3. d. Definition of linear pair
4. b. Linear pair postulate
5. e/m. Definition of supplementary
6. g. Same side interior angles theorem
7. e/m. Definition of supplementary
8. a/c. Substitution property of congruence
9. i. Subtraction property of congruence
10. f. Alternate interior angles theorem
11. l. Alternate exterior angles theorem
12. h. Angle congruence postulate

8 0

2 years ago

Read 2 more answers

What is the correct factorization of x^6-y^6

Soloha48 [4]

(x^3 - y^3)(x^3 + y^3)

You can get this by factoring based on the a^2 - b^2 model

3 0

3 years ago