All categories

3 years ago

Simplify 4x+5/6x^2-x-12 subtract 5x/6x^2-x-12

1 answer:

7 0

4x+5/6x^2-x-12 - (5x/6x^2-x-12)
= (4x + 5 - 5x)/(6x^2-x-12)
= (-x + 5) / (2x - 3)(3x +4)

answer

-x + 5
---------------------
 (2x - 3)(3x +4)

You might be interested in

-5(x+7)<-10 solve for x

olga2289 [7]

-5(x+7)<-10.
You need to use the distributive property.
-5x - 35 < -10.
You need to cancel out -35, so you add it to itself, & -10.
-5x < 25.
You need to divide -5 by itself & 25.
x>-5.

3 0

3 years ago

Read 2 more answers

The graph shows the number of patients seen by a doctor on Tuesday.

MariettaO [177]

B. 4 patients per hour

6 0

3 years ago

What is the decimal multiplier to increase by 93%?

Sidana [21]

Answer:

0.93

Step-by-step explanation:

5 0

2 years ago

4 yards to 30 inches write each ratio in simplest form

MissTica

Answer:

24:5

Step-by-step explanation:

4 yards is 3 feet, so 4*3=12. There are 12 inches in a foot, so 12*12=144. 144:30= 72:15 = 24:5

6 0

2 years ago

A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a s

Umnica [9.8K]

Answer:

88.51 is the minimum score needed to receive a grade of A.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 73

Standard Deviation, σ = 11

We are given that the distribution of exam grades is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.0793.

$P( X \geq x) = P( z \geq \displaystyle\frac{x - 73}{11})=0.0793$

$= 1 -P( z \leq \displaystyle\frac{x - 73}{11})=0.0793$

$=P( z \leq \displaystyle\frac{x - 73}{11})=0.9207$

Calculation the value from standard normal z table, we have,

$P(z \leq 1.410) = 0.9207$

$\displaystyle\frac{x - 73}{11} = 1.410\\x =88.51$

Hence, 88.51 is the minimum score needed to receive a grade of A.

3 0

2 years ago