3^1/5 - 2^5

The Spotlight on Teaching and Learning at the beginning of this chapter discusses the emphasis in the Common Core State Standard

ivanzaharov [21]

I don't know this question

3 0

3 years ago

The price P of a good and the quality Q of a good are linked.

Irina-Kira [14]

the equilibrium point, is when Demand = Supply, namely, when the amount of "Q"uantity demanded by customers is the same as the Quantity supplied by vendors.

That occurs when both of these equations are equal to each other.

let's do away with the denominators, by multiplying both sides by the LCD of all fractions, in this case, 12.

$\bf \stackrel{\textit{Supply}}{-\cfrac{3}{4}Q+35}~~=~~\stackrel{\textit{Demand}}{\cfrac{2}{3}Q+1}\implies \stackrel{\textit{multiplying by 12}}{12\left( -\cfrac{3}{4}Q+35 \right)=12\left( \cfrac{2}{3}Q+1 \right)} \\\\\\ -9Q+420=8Q+12\implies 408=17Q\implies \cfrac{408}{17}=Q\implies \boxed{24=Q} \\\\\\ \stackrel{\textit{using the found Q in the Demand equation}}{P=\cfrac{2}{3}(24)+1}\implies P=16+1\implies \boxed{P=17} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{Equilibrium}{(24,17)}~\hfill$

3 0

3 years ago

Ed plants trees in city parks. He used 1\10 of a bag of fertilizer around the base of a newly planted tree. How many trees can E

valina [46]

To determine this you could think of it in a couple ways.

1. If it takes 1/10 of a bag of fertilizer to put around 1 tree, he can fertilize 10 trees with 1 bag because there are 10 groups of 1/10 in a bag. Take 10 x 40 because there are 40 bags and get 400 trees to be fertilized.

2. You can divide 40 by 1/10 as the other strategy.
This is the same as 40 x 10/1.

Also equaling 400 trees that can be fertilized.

4 0

3 years ago

Identify the domain of the function shown in the graph.

beks73 [17]

The domain of a graph is all of the x-values of the graph. Since the graph looks like it continues and never ends, every single x value is included in the domain, thereby making all reals the correct answer

6 0

3 years ago

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How is the graph of the parent quadratic function transformed to produce the graph of y = negative (2 x + 6) squared + 3?

bearhunter [10]

We start with the parent function

$f(x)=x^2$

The first child function would be

$g(x)=(2x)^2$

We have multiplied the input of the function by a constant: we have

$g(x)=f(2x)$

This kind of transformation result in a horizontal stretch/compression. If the multiplier is greater than 1, we have a compression. So, this first child causes a horizontal compression with compression rate 2.

The second child function would be

$h(x)=(2x+6)^2$

We added 6 to the input of the function: we have

$h(x)=g(x+6)$

This kind of transformation result in a horizontal translation. If the constant added is positive, we translate to the left. So, this second child causes a translation 6 units to the left.

The third child function would be

$l(x)=-(2x+6)^2$

We changed the sign of the previous function (i.e. we multiplied it by -1): we have

$l(x)=-h(x)$

This kind of transformation result in a vertical stretch/compression. If the multiplier is greater than 1 we have a stretch, if it's between 0 and 1 we have compression. If it's negative, we reflect across the x axis, and then apply the stretch/compression. In this case, the multiplier is -1, so we only reflect across the x axis.

The fourth child function would be

$m(x)=-(2x+6)^2+3$

We added 3 to previous function: we have

$m(x)=l(x)+3$

This kind of transformation result in a vertical translation. If the constant added is positive, we translate upwards. So, this last child causes a translation 3 units up.

Recap

Starting from the parent function $y=x^2$ , we have to:

Compress the graph horizontall, with scale factor 2;
Translate the graph 6 units to the left;
Reflect the graph across the x axis;
Translate the graph 3 units up

Note that the order is important!

5 0

3 years ago

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