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Levart [38]
3 years ago
11

What is 76-183+56-1,543 (15 points)

Mathematics
2 answers:
WITCHER [35]3 years ago
7 0
Use Pemdas
<span>76-183+56-1,543
</span><span>-107+56-1,543
</span><span>-51-1,543
</span>−<span>1594
</span>
−1594<span> is your answer.
</span>
Hoped I helped!
san4es73 [151]3 years ago
3 0

Answer: First, I would round all numbers (to the nearest tenth).

183 rounds to 180

76 rounds to 80

15 rounds to 20

29 rounds to 30.

Now plug those numbers into the equation.

180/(80-20)+30

Using PEMDAS, I solve the parentheseis first.

80-20=60.

180/60+30

Then, add 60+30.

60+30=90.

you get 180/90.

180/90=2.

Step-by-step explanation:

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The appropriate response is "0.9476".

Step-by-step explanation:

The given values are:

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As we know,

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On substituting the values, we get

⇒      =\frac{1150}{\sqrt{500} }

then,

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On using the table of P, we get

⇒      =0.9738 - 0.0262

⇒      =0.9476

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is the square root of 81 rational or irrational number and is 5 divided by 3.14 rational or irrational.
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<span>square root of 81 = 9 so it is rational number
 
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</span>
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mart [117]

Answer:

Assuming it is a square with side length 3.2 then the area is:

10.24 units squared

Step-by-step explanation:

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f(x,y)=\dfrac{y^3}x

a. The gradient is

\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath

\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}

b. The gradient at point P(1, 2) is

\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}

c. The derivative of f at P in the direction of \vec u is

D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}

It looks like

\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath

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Then

D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}

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4 years ago
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