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4 years ago

X is all the following except a term a variable a constant an expression

Mathematics

1 answer:

lisov135 [29]4 years ago

6 0

Answer: a constant

Step-by-step explanation:

A constant is a numerical expression like 2, 0,78 etc .

A variable is a alphabetical expression that vary like a,b,c,d,x,y,z.

An expression can consists of both numerical and alphabets and also any arithmetic expression like x, 2abc, 6a+2c etc

A term consist of either numbers and variables multiplied together or only numbers or variable like 2xy, x, 2ab etc.

X is all (a term, a variable , an expression) except a constant because a constant is a numerical expression.

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ra1l [238]

Answer:

42 13/20 km

Step-by-step explanation:

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3 years ago

The temperature at 7 AM was one Fahrenheit the temperature drops 3° every hour for four hours what temperature at 11 AM

castortr0y [4]

Answer:

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Step-by-step explanation:

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3 years ago

1200/180 = (can someone please help me I'm desperate)

ASHA 777 [7]

Using a calculator, 1200/180 = 6.67 approximately

If you use long division, then you'll get what you see in the diagram below

The 6's go on forever after the decimal point, but if you round to two decimal places, then you'll get that approximate value of 6.67

---------------

If you want a remainder, then 1200/180 = 6 remainder 120

You can think of it like you having 1200 cookies and 180 friends. Each friend will get 6 whole cookies and there will be 120 left over as the remainder.

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4 years ago

Consider the differential equation <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D6%20y%5E%7

Nutka1998 [239]

As a Bernoulli equation:

$x^2\dfrac{\mathrm dy}{\mathrm dx}=6y^2+6xy\iff x^2y^{-2}\dfrac{\mathrm dy}{\mathrm dx}-6xy^{-1}=6$

Let $z=y^{-1}\implies\dfrac{\mathrm dz}{\mathrm dx}=-y^{-2}\dfrac{\mathrm dy}{\mathrm dx}$ . The ODE becomes

$-x^2\dfrac{\mathrm dz}{\mathrm dx}-6xz=6$
$x^6\dfrac{\mathrm dz}{\mathrm dx}+6x^5z=-6x^4$
$\dfrac{\mathrm d}{\mathrm dx}[x^6z]=-6x^4$
$x^6z=-6\displaystyle\int x^4\,\mathrm dx$
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$z=-\dfrac6{5x}+\dfrac C{x^6}$
$y^{-1}=-\dfrac6{5x}+\dfrac C{x^6}$
$y=\dfrac1{\frac C{x^6}-\frac6{5x}}$
$y=\dfrac{5x^6}{C-6x^5}$

With $y(3)=6$ , we get

$6=\dfrac{5(3)^6}{C-6(3)^5}\implies C=\dfrac{4131}2$

so the solution is

$y=\dfrac{5x^6}{\frac{4131}2-6x^5}=\dfrac{10x^6}{4131-12x^5}$

which is valid as long as the denominator is not zero, which is the case for all $x\neq\sqrt[5]{\dfrac{4131}{12}}$ .

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3 years ago

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inna [77]

Answer:

Step-by-step explanation:

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3 years ago