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Rzqust [24]
3 years ago
9

100 x 54 + 77 ÷ 2 = ? (RIGHT ANSWER GETS BRAINLIEST)

Mathematics
2 answers:
stealth61 [152]3 years ago
7 0

Answer:

So 2,738.5

Step-by-step explanation:

100×54 equals

— 5,400

5,400+77 equals

—5,477

5,477÷2 equals

—2,738.5

The answer is 2,738,5

Nadusha1986 [10]3 years ago
6 0

Answer:

5438.5

Step-by-step explanation:

To complete this equation, we must use the order of operations also known as PEMDAS.

For the case of this problem, we start by multiplying or dividing numbers before adding or subtracting numbers.

First let's start on the left:

100 x 54 + 77 ÷ 2

5,400 + 77 ÷ 2

5,400 + 38.5

5438.5

I hope this helps! Just remember to always add or subtract number last and start on the left, so that you don't get confused.

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Answer:

  -4u^2/y+(5/2)y^3u^2

Step-by-step explanation:

The applicable rules of exponents are ...

  (a^b)(a^c) = a^(b+c)

  (a^b)/(a^c) = a^(b-c)

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  (8y^2u^7-5y^6u^7)\div(-2y^3u^5)=\dfrac{8y^2u^7-5y^6u^7}{-2y^3u^5}\\\\=-4y^{2-3}u^{7-5}+\dfrac{5}{2}y^{6-3}u^{7-5}=\boxed{\dfrac{-4u^2}{y}+\dfrac{5}{2}y^3u^2}

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2 years ago
Please help with this
mezya [45]

Answer:

96

Step-by-step explanation:

If B is the midpoint of AC, the length of AC has to be twice the length of AB. This is because AB = AC. Twice 48 is 96.

4 0
3 years ago
What is a soliloquy in a drama?
jolli1 [7]
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4 0
3 years ago
Read 2 more answers
Find the horizontal and vertical asymptotes of​ f(x). ​f(x) equals = StartFraction 6 x Over x plus 2 EndFraction 6x x+2 Find the
miss Akunina [59]

Answer:

The horizontal asymptote can be described by the line y = 6

The vertical asymptote can be described by the line x = -2

Step-by-step explanation:

* <em>Lets the meaning of vertical and horizontal asymptotes</em>

- <u><em>Vertical asymptotes</em></u> are vertical lines which correspond to the zeroes

  of the denominator of a rational function

- <u><em>A horizontal asymptote</em></u> is a y-value on a graph which a function

 approaches but does not actually reach

- If the degree of the numerator is less than the degree of the

 denominator, then there is a horizontal asymptote at y = 0

- If the degree of the numerator is greater than the degree of the

 denominator, then there is no horizontal asymptote

- If the degree of the numerator is equal the degree of the denominator,

 then there is a horizontal asymptote at y = leading coefficient of the

 numerator ÷ leading coefficient of the denominator

* <em>Lets solve the problem</em>

∵ f(x)=\frac{6x}{x+2}

∵ The numerator is 6x

∵ The denominator is x + 2

∴ The numerator and the denominator have same degree

∵ The leading coefficient of the numerator is 6

∵ The leading coefficient of the denominator is 1

∴ There is a horizontal asymptote at y = 6/1

∴ <em>The horizontal asymptote can be described by the line y = 6</em>

- Put the denominator equal zero to find its zeroes

∵ The denominator is x + 2

∴ x + 2 = 0

- Subtract 2 from both sides

∴ x = -2

∴ <em>The vertical asymptote can be described by the line x = -2</em>

6 0
3 years ago
Read 2 more answers
If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf
jonny [76]
With the curve C parameterized by

C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k

with 0\le t\le\dfrac\pi2, and given the vector field

\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k

the work done by \mathbf f on a particle moving on along C is given by the line integral

\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt

where

\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt

The integral is then

\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt
=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t
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3 years ago
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