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mart [117]
3 years ago
10

(3x^2+3)+(3x^2+x+4)?

Mathematics
2 answers:
OverLord2011 [107]3 years ago
8 0

Answer:

6x^2+x+7

Step-by-step explanation:

(3x^2+3)+(3x^2+x+4)

6x^2+3+x+4

6x^2+x+7

Alex777 [14]3 years ago
3 0

You first start with adding the number in the parentheses,because none of them can add in this equation it stays the same:

(3x^2 + 3) + (3x ^2 + x + 4)

Since there is nothing that we can do while there are parenthaseese we remove them:

3x^2 + 3 + 3x ^2 + x + 4

Next add the x^2’s:

6x^2 + 3 + x + 4

Lastly the normal numbers:

6x^2 + 7 + x

Now that there are no more numbers that are the same, 6x^2 + x + 7 is your simplified equasion.

Hope this helps! :)

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3 years ago
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krok68 [10]

Answer:

Do not ask for free tutoring (Nobody will really do it)

Step-by-step explanation:

Just use Brainly! Ask for questions and explanations and do what you can to get your work done. After that if you need further help then ask a parent/guardian/sibling or someone you know for help and if that doesn't work then come back to this. Hope this helps.

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3 years ago
PLEASE HELP!!!!!!!!!!!!! DUE SOON
Sergio039 [100]

\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h=-16t^2+\stackrel{\stackrel{v_o}{\downarrow }}{65}t


now, take a look at the picture below, so for 2) and 3) is the vertex of this quadratic equation, 2) is the y-coordinate and 3) the x-coordinate.


\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+65}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{65}{2(-16)}~~,~~0-\cfrac{65^2}{4(-16)} \right) \implies \left( \cfrac{65}{32}~,~0- \cfrac{4225}{-64}\right)


\bf \left( \cfrac{65}{32}~,~0+ \cfrac{4225}{64}\right)\implies \left( \stackrel{seconds}{2\frac{1}{32}}~~,~~ \stackrel{feet~hight}{66\frac{1}{64}}\right)

6 0
3 years ago
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Alexxx [7]

Answer:

2.5

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
PLEASE HELP!!!!!!!
lozanna [386]
The correct answer is: <span>C) The base of the cone and the top of the cylinder have the same area. </span>The cone has the smallest volume of the 2 figures. This is because the formula for the cylinder is b x h, the formula for the cone is 1/3(b x h) so if they have same height and base area cylinder would have larger volume because, for the cylinder, a formula is one-third of b x h. Hope I helped!!  : )

hope this helps you


4 0
3 years ago
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