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2 years ago

6 1/4 - 3 5/8 please help im too lazy to do all the work. I WILL GIVE 20 POINTS!

Mathematics

1 answer:

IRINA_888 [86]2 years ago

8 0

Answer:

2 5/8

Step-by-step explanation:

simple met... I mean math :)

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How can you use distributive property to find 3/4 times 2 and 1/3

zubka84 [21]

Answer:

1 3/4

Step-by-step explanation:

You can set it up like the distrubtive property first

3/4(2+ 1/3) Next you use the distirbutive property instead of adding inside the parenthesis and you multiply the 3/4 by 2 and then the 3/4 by 1/3

Use 2 as 2/1 in fraction problems

3/4 * 2/1 Now multiply the numerators together and the denomenators together

3 * 2 = 6 4 * 1 =4

6/4

3/4 *1/3

3 * 1 =3

4 * 3 =12

3/12

Now we have to make them both like terms (same denominator) so we can add them, we will multiply the 6/4 by 3 (Both top and bottom)

6 * 3 = 18 4 *3 =12

18/12

Now we add the fractions

3/12 +18/12 = 21/12

Reduce that and you get

21/12 = 1 9/12 or 1 3/4

5 0

3 years ago

ANSWER TO BE NAMED BRAINLIEST GET A THANK YOU AND A 5 STAR VOTE FOR YOUR ANSWER PLZ HELP

tensa zangetsu [6.8K]

Answer:

80, every 5 percent is 4 people. So every 25% is 20. 20 x 4 is 80.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Directions: Find the indicated trigonometric ratio as a fraction in simplest form. 1. Sin L =

Rainbow [258]

Answer:

$\sin L = 0.60$

$tan\ N = 1.33$

$\cos L = 0.80$

$\sin N = 0.80$

Step-by-step explanation:

Given

See attachment

From the attachment, we have:

$MN = 6$

$LN = 10$

First, we need to calculate length LM,

Using Pythagoras theorem:

$LN^2 = MN^2 + LM^2$

$10^2 = 6^2 + LM^2$

$100 = 36 + LM^2$

Collect Like Terms

$LM^2 = 100 - 36$

$LM^2 = 64$

$LM = 8$

Solving (a): $\sin L$

$\sin L = \frac{Opposite}{Hypotenuse}$

$\sin L = \frac{MN}{LN}$

Substitute values for MN and LN

$\sin L = \frac{6}{10}$

$\sin L = 0.60$

Solving (b): $tan\ N$

$tan\ N = \frac{Opposite}{Adjacent}$

$tan\ N = \frac{LM}{MN}$

Substitute values for LM and MN

$tan\ N = \frac{8}{6}$

$tan\ N = 1.33$

Solving (c): $\cos L$

$\cos L = \frac{Adjacent}{Hypotenuse}$

$\cos L = \frac{LM}{LN}$

Substitute values for LN and LM

$\cos L = \frac{8}{10}$

$\cos L = 0.80$

Solving (d): $\sin N$

$\sin N = \frac{Opposite}{Hypotenuse}$

$\sin N = \frac{LM}{LN}$

Substitute values for LM and LN

$\sin N = \frac{8}{10}$

$\sin N = 0.80$

7 0

3 years ago

For questions 1 and 2, use factoring to find the x-intercepts of these quadratic equations.

Marysya12 [62]

1.x^2 +6x - 4 = 6x
X^2 + 6x -4 - 6x = 0
X^2 - 4 = 0
By difference of two squares :
X^2 - 4 = 0 can be written as (x-2) (x+2) = 0
(X-2)=0 therefore x= 2

(X+2)=0 therefore x= -2

2. X^2 - 8x = -6x
X^2 -8x + 6x = 0
X^2 - 2x = 0
X(X-2) = 0
X= 0, X= 2

6 0

3 years ago

Which graph shows a function whose inverse is also a function?

strojnjashka [21]

Answer:

Step-by-step explanation:

edge2021

3 0

3 years ago