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SpyIntel [72]
3 years ago
8

99 POINTS

Mathematics
1 answer:
yan [13]3 years ago
3 0

Answer:

the function given f(x)=x^3

for 4 units up, just add 4 as a constant

4 units up means, every old value will now be 4 more than previous value. at x=0, y=0 in the transformed curse it should x=0 and y=4, so just add it.

f(x)=x^3+4

for 6 units left,

each old value of y should now occur 6 units before the old value of x i.e. X=x+6

for example, the point (0,0) should occur at (-6,0) in the transformed graph,

hence, f(x)=(x+6)^3

so the final curve is

f(x)=(x+6)^3+4

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suppose your restaurant bill comes to $28.35. Find your total cost if the tax 6.25% and you leave a 20% tip on the amount before
Elden [556K]
36.34 dollars    28.50*1.625=30.28*1.2=36.34

6 0
3 years ago
I dont know how to do this
ra1l [238]
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3 0
3 years ago
Ava has $110 in her piggy bank and spends $5 a week. Liam has $20 in his piggy bank and does chores to earn an extra $4 a week.
jarptica [38.1K]

Answer:

Ava and Liam have the same amount of money at 10th week.

Step-by-step explanation:

Given:

Amount Ava has in her piggy bank = $110

Amount spends per week = $5

Let the number of weeks be 'w'.

So we can say that.

Total amount Ava had after 'w' weeks = 110-5w

Also Given:

Amount Liam has in her piggy bank = $20

Amount earns per week = $4

Let the number of weeks be 'w'.

So we can say that.

Total amount Liam had after 'w' weeks = 20+4w

Now we need to find the number of weeks when both will have same amount.

So we can say that;

Total amount Ava had after 'w' weeks =Total amount Liam had after 'w' weeks

110-5w=20+4w

Combining like terms we get;

110-20=5w+4w\\\\90=9w

Dividing both side by 9 we get;

\frac{90}{9}=\frac{9w}{9}\\\\w=10\ weeks

Hence Ava and Liam have the same amount of money at 10th week.

8 0
4 years ago
Is -2.2675 a irrational number?
bezimeni [28]
No because it has an ending it doesnt continue 

5 0
3 years ago
How do you do this problem? I need to know how you found the answer.
Alexxandr [17]

to get the equation of a line, we simply need two points, say for the Red one ... notice in the graph the lines passes through (0,2) and (-1,6), so let's use those


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{-1-0}\implies \cfrac{4}{-1}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=-4(x-0) \\\\\\ y-2=-4x\implies \blacktriangleright y=-4x+2 \blacktriangleleft


now, for the Blue one, say let's use hmmm it passes through (0,2) and (1.6)


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{1-0}\implies \cfrac{4}{1}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=4(x-0) \\\\\\ y-2=4x\implies \blacktriangleright y=4x+2 \blacktriangleleft

3 0
3 years ago
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