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3 years ago

Is 5/7 irrational number

Mathematics

2 answers:

amm18123 years ago

8 0

Answer:

5/7 is a rational number. The word 'rational' actually comes from the word 'ratio. ' One way to write a ratio is with a fraction.

Step-by-step explanation:

Alex73 [517]3 years ago

5 0

Answer:yes

Step-by-step explanation:

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A student scored 83 and 91 on her first two quizzes. Write and solve a compound inequality to find the possible values for a thi

mixer [17]

Let x be the score of the 3rd quiz,, then the average score will be:
(83+91+x)/3, which = (174+x)/3. but this average should be greater or equal to 85 and smaller or equal to 90:

85 ≤ (174+x)/3 ≤ 90
1st multiply all terms with 3 → 255 ≤(174+x) ≤ 270
2nd subtract 173 from all terms → 255-174 ≤174-174+x ≤270-174
81 ≤ x ≤ 96

7 0

3 years ago

Is 19,279 divisibility by 2

erastovalidia [21]

Answer:

it isn't

Step-by-step explanation:

it isn't because the last digit isn't divisible by 2

6 0

3 years ago

Read 2 more answers

I need help with this one can anyone help ?

Novosadov [1.4K]

(F) 0.2

10/50 = 1/5 = 0.2

(G) 1.6667

5/3 = 1 + (2/3) = 1 + 0.666....

(H) 1.11394

Use a calculator

(I) DNE

Dividing by 0 is not allowed.

6 0

2 years ago

The perimeter of a triangle is 60 feet. one side is 12 feet long. of the two unknown sides, one of them is twice as long as the

Lady bird [3.3K]

One is 36 the other is 12

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3 years ago

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