All categories

3 years ago

F(x)=1/4x^2+1/2x-35/4 how do I factor fractions?

1 answer:

5 0

let's multiply both sides by the LCD of all fractions, in this case is 4, just to do away with the denominators for a few seconds

$\bf f(x) = \cfrac{1}{4}x^2+\cfrac{1}{2}x-\cfrac{35}{4}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{4}}{4[f(x)]=4\left( \cfrac{1}{4}x^2+\cfrac{1}{2}x-\cfrac{35}{4} \right)} \\\\\\ 4f(x) = x^2+2x-35\implies 4f(x) = (x+7)(x-5) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill f(x) = \cfrac{1}{4}(x+7)(x-5)~\hfill$

You might be interested in

How to find the common denominator of two numbers?

Snowcat [4.5K]

Find the number that both can be divided by i.e. the common denominator of 56/8 is 8 because 56 is divisible by 8 and 8 is also divisible by 8.
so by finding the common denominator and dividing the fraction, the equation would look like this: 7/1

3 0

3 years ago

A professor grades students on four tests, a term paper, and a final examination. Each test counts as 15% of the course grade. T

alina1380 [7]

Answer:

The Alans's average for the course is 84.5

Step-by-step explanation:

We are given

There are 4 tests, A Term paper and a Final examination.

Score = 92, 78, 82, 90.

Term paper score = 80

Final examination score = 86

Weighted mean = ∑ w.f/∑w

also the sum of all the weights is 100% = 1

weighted mean = 15%*92 + 15%*78+15%*82+15%*90+20%*80+20%*86/1

= 51.3 + 33.2

= 84.5

Therefore the Alans's average for the course is 84.5

5 0

2 years ago

2p +4p=20, in a verbal sentence

maria [59]

Answer:

see below

Step-by-step explanation:

Twice p added to 4 multiplied by p is equal to 20

3 0

3 years ago

Which of the following is not an example of replication?

Vitek1552 [10]

Solution: Which of the following is not an example of replication?

Answer: Replication is the repetition of an experimental condition so that the variability associated with the phenomenon can be estimated. Therefore option A is not an example of replication

A. Conducting different experiments on the same groups

3 0

3 years ago

Determine whether the set of vectors <img src="https://tex.z-dn.net/?f=%20v_%7B1%3D%283%2C2%2C1%29%2C%20v_%7B2%7D%20%3D%28-1%2C-

Korolek [52]

Since each vector is a member of $\mathbb R^3$ , the vectors will span $\mathbb R^3$ if they form a basis for $\mathbb R^3$ , which requires that they be linearly independent of one another.

To show this, you have to establish that the only linear combination of the three vectors $c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3$ that gives the zero vector $\mathbf0$ occurs for scalars $c_1=c_2=c_3=0$ .

$c_1\begin{bmatrix}3\\2\\1\end{bmatrix}+c_2\begin{bmatrix}-1\\-2\\-4\end{bmatrix}+c_3\begin{bmatrix}1\\1\\-1\end{bmatrix}=(0,0,0)\iff\begin{bmatrix}3&-1&1\\2&-2&1\\1&-4&-1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$

Solving this, you'll find that $c_1=c_2=c_3=0$ , so the vectors are indeed linearly independent, thus forming a basis for $\mathbb R^3$ and therefore they must span $\mathbb R^3$ .

4 0

3 years ago