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3 years ago

The quotient of 45 and 2 6/7

1 answer:

8 0

To find the quotient of 45 and 3 6/7,
(45) / (3 6/7)
(45) / (27/7) [convert the mixed number to an improper fraction]
(45) * (7/27) [to divide by fraction, invert denominator and multiply]
315 / 27
35 / 3 [optionally, reduce fraction]
11 2/3 [convert improper fraction to mixed number]

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36÷(6÷3) plz helppppooooop

Sunny_sXe [5.5K]

Answer:

Step-by-step explanation:

Divide with by paratheses first. Use PEMDAS method

7 0

2 years ago

Read 2 more answers

1.2million users of social media via mobile devices were added each day. How many users added on an average each second

Pachacha [2.7K]

Answer: 14 new users are added per second

Step-by-step explanation:

1,200,000/24 = 50000

50000/60 = 833.333333333

833.333333333/60 = 13.8888888889

then round because it said on average

so if we round 13.8888888889 we get 14

so on average 14 new users are added per second.

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3 years ago

Ok I rlly need help with this question it’s rlly hard for me there’s a picture to show —> helppp

Whitepunk [10]

Answer:

x=20; vertical angles theorem

Step-by-step explanation:

hi! we can use the vertical angles theorem for this. vertical angles are congruent to each other, meaning they are the same amount of degrees. so, we can set 2x+10 and 50 equal to each other.

2x+10=50

2x=40

x=20

x=20 because of the vertical angles theorem.

8 0

2 years ago

I have $1500 to invest in an account that pays 1.25% compounded continuously. How much money will I have after 10 years?

S_A_V [24]

I hope you find my answer helpful I got 39,000.

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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