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2 years ago

Someone, please answer. I've been asking this question for TWO DAYS NOW!

Mathematics

1 answer:

Leni [432]2 years ago

3 0

So I labeled it as (1,7) as A, (-5,7) as B, and (-5,3) as C.So A is (0,5) and B as (-3,5) and then C as (-3,3).I think that would be correct.

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A bakery has the following pricing for large orders of cupcakes. The first 100 cupcakes of any order

Taya2010 [7]

A piecewise function is a function that behaves differently in different intervals of its domain.

The answers are:

f(x) = x*$2.00 if 0 ≤ x ≤ 100

f(x) = x*$1.75 if 100 < x ≤ 250

f(x) = x*$1.25 if x > 250.

b) $306.25

The given information is:

The first 100 cupcakes cost $2.00 each.

If the order is between 100 and 250 cupcakes, the cost is $1.75 each.

If the order is more than 250 cupcakes, the cost is $1.25 each.

a) Now we want to write the piecewise function, this will be:

f(x) = x*$2.00 if 0 ≤ x ≤ 100

f(x) = x*$1.75 if 100 < x ≤ 250

f(x) = x*$1.25 if x > 250.

Above you can see the piecewise function, when you want to evaluate it, the first thing you need to do is to find at which interval the input belongs.

b) If you order 175 cupcakes you have x = 175.

This belongs to the second interval ( 100 < x ≤ 250) then we will use the second part of the function:

f(175) = 175*$1.75 = $306.25

If you want to learn more, you can read:

brainly.com/question/12561612

8 0

2 years ago

The side length of the base of a square pyramid is 30 centimeters. The height of the pyramid is 45 centimeters.

puteri [66]

Answer:

volume = 13500 $cm^{3}$

Step-by-step explanation:

For a given pyramid, its volume can be determined by:

volume of pyramid = $\frac{lwh}{3}$

Where: l is the base length, w is the base width and h id the height of the pyramid.

For the given question, l 30 cm, w = 30 cm and h = 45 cm.

So that,

volume = $\frac{30*30*45}{3}$

= $\frac{40500}{3}$

= 13500

volume = 13500 $cm^{3}$

The volume of the pyramid is 13500 $cm^{3}$ .

6 0

2 years ago

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .