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3 years ago

5x-1/5=7/3 Find The LCD That would eliminate the fractions

Mathematics

1 answer:

photoshop1234 [79]3 years ago

7 0

Answer:

$x=\frac{7}{15}$

Step-by-step explanation:

$5x-\frac{1}{5} =\frac{7}{3}$

$5x - \frac{3}{15} = \frac{35}{15}$

$5x - \frac{3}{15} + \frac{3}{15} = \frac{35}{15} + \frac{3}{15}$

$5x=\frac{35}{15}$

$\frac{5x}{5} } =\frac{\frac{35}{15}}{5}$

$x=\frac{7}{15}$

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There are 38 students. If there are 6 seats at each table, how many tables would be needed for all of the students to be able to

Leto [7]

Answer:

7 tables

Step-by-step explanation:

So if 6 students can fit at each table, you divide 38 by 6. You get 6 1/3 ,but since you can't get 1/3 of a table, you have to get 7

4 0

3 years ago

Ellen flipped a coin 80 times. the coin landed heads up 44 times and tails up 36 times. what is the experimental probability of

Vaselesa [24]

Answer:

$\frac{9}{20}$

Step-by-step explanation:

experimental probability is the probability of an event occuring when we perform an experiment.

It goes according to the results of the experiment, not the theory behind it.

<u>Probability of landing tails up, therefore, is the number of times it came tails up divided by the total number of times the coin was flipped.</u>

<u />

Hence

P (tails) = 36/80 = 9/20

5 0

3 years ago

Read 2 more answers

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2