Ask question

Ask question

All categories

2 years ago

8

Solving rational equations solve for x: 3/x-4=7/x

2 answers:

KIM [24]2 years ago

8 0

Solving the rational equation $\frac{3}{x-4}=\frac{7}{x}$ the value of x is 7

Step-by-step explanation:

We need to solve rational equation: $\frac{3}{x-4}=\frac{7}{x}$ and find value of x.

Solving:

$\frac{3}{x-4}=\frac{7}{x}$

Cross multiplying:

$3(x)=7(x-4)\\Simplifying:\\3x=7x-28\\3x-7x=-28\\-4x=-28\\x=-28/-4\\x=7$

So, value of x = 7.

Solving the rational equation $\frac{3}{x-4}=\frac{7}{x}$ the value of x is 7

Keywords: Solving the rational equation

Learn more about Solving the rational equation at:

#learnwithBrainly

Olenka [21]2 years ago

4 0

Answer:

-1

Step-by-step explanation:

3/x - 4 = 7/x

3/x - 7/x = 4

-4/x = 4

x = -1

You might be interested in

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

3 years ago

One side of a square is 8 centimeters long what is the area

Leni [432]

A= b(h)
a= 8(8)
a= 64cm^2

5 0

2 years ago

Apple weights are normally distriuted with a mean 0.33 pounds and and standard deviation of 0.06 pounds. Jermaine went to the st

maksim [4K]

Using the normal distribution, it is found that 10.56% of the apples weigh less than Jermaine's apple.

--------------------

In a <em>normal distribution</em> with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure X is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X, that is, the proportion of measures that are less than X.

In this problem:

Mean of 0.33 pounds, thus $\mu = 0.33$ .
Standard deviation of 0.06 pounds, thus $\sigma = 0.06$ .
The proportion that is less than 0.25 pounds is the p-value of Z when X = 0.25, thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.25 - 0.33}{0.06}$

$Z = -1.25$

$Z = -1.25$ has a p-value of 0.1056.

0.1056 x 100% = 10.56%.

10.56% of the apples weigh less than Jermaine's apple.

A similar problem is given at brainly.com/question/13411796

8 0

2 years ago

Please give me explanation please please please solve it

insens350 [35]

i hope this was helpful!

5 0

2 years ago

What is the biggest area of a quadrilateral you can fit in a circle that has a

erik [133]

Answer: 200 square units

Step-by-step explanation:

Given

Circumference of the circle is $20\pi$

Suppose r is the radius of the circle

The biggest area of a quadrilateral that can fit in a circle is of square.

Deduce the radius of the circle

$2\pi r=20\pi\\r=10\ \text{units}$

Suppose the side of the square is a

from the figure, we can write

$\Rightarrow a^2+a^2=(2r)^2\\\Rightarrow 2a^2=4r^2\\\Rightarrow a=\sqrt{2}r\\\Rightarrow a=10\sqrt{2}\ \text{units}$

Area of the quadrilateral is

$\Rightarrow A=(10\sqrt{2})^2\\\Rightarrow A=100\times 2\\\Rightarrow A=200\ \text{square units}$

7 0

3 years ago