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k0ka [10]
3 years ago
7

What number, when subtracted by 15, equals 18?

Mathematics
2 answers:
jok3333 [9.3K]3 years ago
8 0

Answer:

33

Step-by-step explanation:

Because 15+18=33. If you subtract 15 from 33 , you will get 18

Therefore, my answer is correct.

ad-work [718]3 years ago
7 0

Answer:

23

Step-by-step explanation:

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Given the rule, find the 8th term in the arithmetic<br> sequence.<br> an =3n + 4
dybincka [34]

Answer:

You simply substitute 8 for n

A₈ =3(8)+4=28

4 0
2 years ago
Prime numbers have
sp2606 [1]
Exactly two factors
3 0
3 years ago
What is -6x+4=-50 and 5x+6=-44 also -3(7+6x)=-201 please solve these equations and show work
grandymaker [24]
Remember you can do anyting to an equation as long as ou do it to both sides



-6x+4=-50
minus 4 both sides
-6x=-54
divide both sides by -6
x=9



5x+6=-44
minus 6 both sides
5x=-50
divide both sides by 5
x=-10


-3(7+6x)=-201
normally you would distribute (that works) but it's easier to divide both sides by -3
7+6x=67
minus 7 both sides
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divide by 6 both sides
x=10
6 0
3 years ago
Explain why each non-zero integer has two square roots but only one cube root.
lilavasa [31]

if we have a number like say hmm 4, and we say hmmm √4 is ±2, it simply means, that if we multiply that number twice by itself, we get what's inside the root, we get the 4, so (+2)(+2) = 4, and (-2)(-2) = 4, recall that <u>minus times minus = plus</u>.

so, any when we're referring to even roots like \bf \sqrt[2]{~~},\sqrt[4]{~~},\sqrt[6]{~~}...., the positive number, that can multiply itself an even amount of times, will produce a valid value, BUT the negative number that multiply itself an even amount of times, will also produce a valid value.

now, that's is not true for odd roots like \bf \sqrt[3]{~~},\sqrt[5]{~~},\sqrt[7]{~~}...., because the multiplication of the negative number will not produce a valid value, let's put two examples on that.


\bf \sqrt[3]{27}\implies \sqrt[3]{3^3}\implies 3\qquad because\qquad (3)(3)(3)=27&#10;\\\\\\&#10;however\qquad (-3)(-3)(-3)\ne 27~\hspace{8em}(-3)(-3)(-3)=-27&#10;\\\\[-0.35em]&#10;\rule{34em}{0.25pt}\\\\&#10;\sqrt[3]{-125}\implies \sqrt[3]{-5^3}\implies -5\qquad because\qquad (-5)(-5)(-5)=-125&#10;\\\\\\&#10;however\qquad (5)(5)(5)\ne -125~\hspace{10em}(5)(5)(5)=125


so, when the root is an odd root, you will always get only one number that will produce the radicand.

6 0
3 years ago
One half of the sum of 7 and F minus the product of 6 and Q?
ankoles [38]
This is how you would write the statement:
\frac{1}{2}(7+F)-(6*Q)

You could also write it like this, if the one-half applies to everything:
\frac{1}{2} ((7+F)-(6*Q))
8 0
3 years ago
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