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Tom [10]
2 years ago
15

Joseph has $60 in a savings account. Each week he deposits $12 in the account.

Mathematics
1 answer:
Aleonysh [2.5K]2 years ago
5 0

Answer:

12x +60

Step-by-step explanation:

c

d

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A cube has a side length of 6 centimeters what is the volume
Semenov [28]

Answer:

the answer is 24

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3 years ago
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Which zero pair could be added to the function fon) = x2 + 12x + 6 so that the function can be written in vertex form?
Oksanka [162]

ANSWER

36,-36

EXPLANATION

The given function is:

f(x) =  {x}^{2}  + 12x + 6

To write this function in vertex form;

We need to add and subtract the square of half the coefficient of x.

The coefficient of x is 12.

Half of it is 6.

The square of 6 is 36.

Therefore we add and subtract 36.

Hence the zero pair is:

36, -36.

The correct answer is D.

4 0
3 years ago
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Factorise y^2-10y+16
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I hope this helps you

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3 years ago
Question 1 Part A One of Maria and Josie’s early tasks is printing and laminating signs to hang at local businesses around town.
maks197457 [2]

Answer:

Using the table

Maria     does 1/4 of the job in 1 hour since it takes her 4 hours to complete the job  , she works x hours and completes x/4 of the task

Josie     does 1/2 of the job in 1 hour since it takes her 2 hours to complete the job  , she works x hours and completes x/2

x/4 + x/2 = 1 completed task

Multiply by 4

x/4 *4 + x/2*4 = 1*4

x+2x =4

3x=4

x = 4/3

x = 1 1/3 hours

5 0
3 years ago
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Hello can someone help me simplify 4) and 5)? Thank you and please include steps :)
Leni [432]
Alrighty

remember some rules
(ab)^c=(a^c)(b^c)
and
x^{-m}=\frac{1}{x^m}
and
x^0=1 for all real values of x
and
(a^b)^c=a^{bc}
and
(\frac{a}{b})^c=\frac{a^c}{b^c}
and
(a/b)/(c/d)=(ad)/(bc)
and
(a^b)(a^c)=a^{b+c}
and
\frac{a^m}{a^n}=a^{m-n}
and don't forget pemdas
example: -x^m=-1(x^m) but (-x)^m=(-1)^m(x^m)

so

4.
(\frac{c^{-2}}{2})^{-2}=

\frac{(c^{-2})^{-2}}{2^{-2}}=

\frac{c^4}{\frac{1}{2^2}}=

\frac{c^4}{\frac{1}{4}}=

4c^4



5.

\frac{(-a)^4bc^5}{-a^2b^{-3}c^0}=

\frac{(-1)^4(a)^4bc^5}{-1(a^2)(\frac{1}{b^3}(1)}=

\frac{(1)a^4bc^5}{\frac{(-1)(a^2)}{b^3}}=

\frac{(a^4bc^5)b^3}{(-1)(a^2)}=

\frac{-a^4b^4c^5}{a^2}=

-a^2b^4c^5
3 0
3 years ago
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