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3 years ago

How to solve 4x / 2 = -5

Mathematics

2 answers:

Butoxors [25]3 years ago

7 0

Answer:

-5/2

Step-by-step explanation:

4x/2 = -5

We need to isolate the x. In order to do that we'd first multiply both sides by two. That leaves us with:

4x = -10

Next we need to divide both sides by four to get rid of the four. That leaves us with:

x = -10/4

Finally, we'd need to simplify our answer. That leaves us with a final answer of:

-5/2

Hope this helps!! <3

patriot [66]3 years ago

6 0

Answer:

Step-by-step explanation:

4x/2= -5

4x= -5*2

2x= -5

x= -5/2

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victus00 [196]

Given that a coupon for $15 off will be applied on purchases over $100, the given discounted price of the jacket is $203. Since this is the discounted price, then it should be the price after the coupon has been used. To find the retail (or original) price, we simply add 15 back to 203. So the retail price should be $218.

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2 years ago

Read 2 more answers

14, 15 and 16 please

WARRIOR [948]

14). (Take π value as 22/7)
circumference = 66in
2πr = 66in
2×22/7×r = 66
r = 66×7/2×22
r = 21/2in
d = 2r => d = 2×21/2
therefore diameter of the circle = 21in

15). (Take π value as 3.14)
Circumference = 3.14m
2πr = 3.14m
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r = 1/2m
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16). (Take π value as 22/7)
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2πr = 33km
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d = 2×21/4
d = 21/2km
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3 years ago

the sum of two numbers is 49.Twice the smaller number is 97 less than 3 times the larger number.What are the numbers. help mee I

kondaur [170]

Answer:

smaller number is 10

larger number is 39

Step-by-step explanation:

a = small number

b = larger number

a + b = 49

2a = 3b - 97

substitute 49-a into second equation for b

2a = 3(49-a) - 97

2a = 147 - 3a - 97

5a = 50

a = 10

b = 39

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2 years ago

Tolong di jawab pertanyaan ini ya no 1 sampai 6

umka21 [38]

Que ldixldikxhskzjskjzss

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2 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

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2 years ago