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3 years ago

Name the sets of numbers to which 7/9 belongs

Mathematics

1 answer:

tia_tia [17]3 years ago

6 0

Answer:

OK.

the sets are

natural numbers (or counting numbers), this is like 1,2,3,4,5 etc

whole numbers, this is including 0, so 0,1,2,3,4,5,6 etc

integers, this includes the previoius set and negatives, so -3,-2,-1,0,1,2,3,4,5 etc

rational numbers, this is the set of numbers that can be written in form a/b where b≠0, so all integers can be written like this, like example -3=-3/1, so -7/9 belongs here

-7/9 goes in the rational set

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Solve for E: <br> E=〖mc〗^2<br> m = 3<br> c = 6

rewona [7]

Answer:

E = 324

Step by step explanation:

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3 years ago

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(01.04 MC)Simplify the following expression: (−4)(−2)(−3) A -24 B -11 C 11 D 24

lys-0071 [83]

Answer:

-24

Step-by-step explanation:

So in short this is what it means:

-4×-2×-3

So to work it out this is what you do:

Deal with the first two numbers; -4 × -2=8

It becomes an eight without a negative because when multiplying numbers with the same signs(positive or negative) the answer is always a positive number thus the answer becomes 8.

Then 8×-3=-24

This is because when multiplying a negative and positive numbers the answer is always a negative numbers thus -24.

So automatically the answer is -24.

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3 years ago

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Which equation has infinite solutions?

AnnyKZ [126]

The equation that has an infinite number of solutions is $2x + 3 = \frac{1}{2}(4x + 2) + 2$

<h3>How to determine the equation?</h3>

An equation that has an infinite number of solutions would be in the form

a = a

This means that both sides of the equation would be the same

Start by simplifying the options

3(x – 1) = x + 2(x + 1) + 1

3x - 3 = x + 3x + 2 + 1

3x - 3 = 4x + 3

Evaluate

x = 6 ----- one solution

x – 4(x + 1) = –3(x + 1) + 1

x - 4x - 4 = -3x - 3 + 1

-3x - 4 = -3x - 2

-4 = -2 ---- no solution

$2x + 3 = \frac{1}{2}(4x + 2) + 2$

2x + 3 = 2x + 1 + 2

2x + 3 = 2x + 3

Subtract 2x

3 = 3 ---- infinite solution

Hence, the equation that has an infinite number of solutions is $2x + 3 = \frac{1}{2}(4x + 2) + 2$

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<u>Complete question</u>

Which equation has infinite solutions?

3(x – 1) = x + 2(x + 1) + 1

x – 4(x + 1) = –3(x + 1) + 1

$2x + 3 = \frac{1}{2}(4x + 2) + 2$

$\frac 13(6x - 3) = 3(x + 1) - x - 2$

5 0

2 years ago

Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

3 years ago

PLEAES HELP ME I SUCK AT MATH!!

ehidna [41]

Answer:

its 5

Step-by-step explanation:

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3 years ago