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slava [35]
3 years ago
14

Order these numbers from least to greatest. 154 , 3911 , 3.735 , 3.79

Mathematics
2 answers:
kotegsom [21]3 years ago
8 0
3.735, 3.79, 154, 3911
Pachacha [2.7K]3 years ago
4 0
3.735<3.79<154<3911....................
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CRS-2 emulsion weighs 8.4lb/gal. I purchased 1.51Ton. How many gallons did I purchase?
disa [49]

2000 pounds = 1 ton

1.51 x 2000 = 3020 pounds

3020/8.4 = 359.52 gallons. Round off answer as needed.

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3 years ago
Use f’( x ) = lim With h ---&gt; 0 [f( x + h ) - f ( x )]/h to find the derivative at x for the given function. 5-x²
beks73 [17]
<h2>Answer:</h2>

The derivative of the function f(x) is:

                 f'(x)=-2x

<h2>Step-by-step explanation:</h2>

We are given a function f(x) as:

f(x)=5-x^2

We have:

f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)

( Since,

(a+b)^2=a^2+b^2+2ab )

Hence, we get:

f(x+h)=5-x^2-h^2-2xh

Also, by using the definition of f'(x) i.e.

f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}

Hence, on putting the value in the formula:

f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x

      Hence, the derivative of the function f(x) is:

          f'(x)=-2x

3 0
3 years ago
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Greeley [361]

workings...

3x4=12

answer: 12/5 or 2 2/5

4 0
4 years ago
Steak cost $3.85 per pound if you buy a 3 pound steak and pay for it with a $20 bill how much change will you get ?
Pie

Answer:

$ 8.45

Step-by-step explanation:

3.85 * 3 = 11.55

20 - 11.55 = 8.45

3 0
3 years ago
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Help please , i just don’t understand pre cal
Masja [62]

Answer:

In mathematics education, precalculus is a course, or a set of courses, that includes algebra and trigonometry at a level which is designed to prepare students for the study of calculus. Schools often distinguish between algebra and trigonometry as two separate parts of the coursework.Step-by-step explanation:

hope this help

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3 years ago
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