What is 3(2x-1)-2(x-5) Please it's urgent and I forgot how to do it

Mathematics

1 answer:

RideAnS [48]3 years ago

4 0

The answer for 3(2x-1)-2(x-5) is 4x+7

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Which equations are equivalent to the given equation?

Colt1911 [192]

The answer is B
X-5= 11.2
you want to get x by itself so you add 5 on both sides
X= 11.2+ 5
Hope this helps!!!

3 0

4 years ago

tom had a platter of chocolate wafers. he ate 5 of them and then gave his brother 3 he then moved on to his baseball team of 8 m

Aloiza [94]

Tom started with total 72 chocolate wafers.

Explanation

The number of chocolate wafers taken by 8 members of the baseball team are in the sequence : $1, 3,5,7,9,11,13,15$

The above sequence is arithmetic sequence with first term(a₁)= 1 and common difference (d) = 2

Formula for Sum of first $n$ terms in arithmetic sequence is....

$S_{n}= \frac{n}{2}[2a_{1}+(n-1)d]$

So, the Sum of 8 terms in that sequence....

$S_{8}= \frac{8}{2}[2(1)+(8-1)(2)]\\ \\ S_{8}= 4[2+7(2)]\\ \\ S_{8}=4(2+14)\\ \\ S_{8}=4(16)=64$

That means, the total number of chocolate wafers taken by the baseball team members is 64. Tom ate 5 and then gave his brother 3 chocolate wafers at first.

So, the total number of chocolate wafers at starting $=64+5+3=72$

6 0

4 years ago

A rectangular prism has a length of 20 in., a width of 2 in., and a height of 314 in.

Novay_Z [31]

First solve the volume of the prism
V = l x w x h
where l is the length
w is the width
h is the hieght

V = 20 x 2 x 314
V = 12560 cu in

then solve the volume of the cube
V = e^3
V = 14^3
V = 2774 cu in

number of cube = 12560 / 2774
number of cube = 4 cubes

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3 years ago

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ASHA 777 [7]

$\left( 16 x^8 y^{-12} \right)^{\frac 1 2} = 16^{\frac 1 2} (x^8)^{\frac 1 2} (y^{-12})^{\frac 1 2} = 4 x^4 y^{-6} = \dfrac{4 x^4}{y^6}$

Third choice

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3 years ago

What is a good acronym to remember the metric conversions?

Elenna [48]

I use the acronym
Giga-Gosh
Mega-Molly
Kilo-Kathryn
Hecto-Hates
Deca-Dad
deci-dad
centi-can't
mili-multiply
micro-many
nano-numbers

I hope this helps, please let me know if you have any other questions or need any other help! Thank you! :)

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3 years ago