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3 years ago

Use the table from Callie's Candies to answer the question.

Mathematics

1 answer:

frez [133]3 years ago

6 0

We can use logic along with 3 linear equations to solve this problem.

For the three types of candies, we will write a slope-intercept form equation. We know what m (slope) is for each equation, and there is no y-intercept because there is no starting point.

Equations:

Mints: y=.96x
Chocolates: y=4.70x
Lollipops: y=.07x

Using the given information, we can use the equations in function form. We know what x (input) is for all three types of candy, and that will give us y (output), which is the total for that candy type.

Solving:

Mints: y=.96(.75)
Chocolates: y=4.70(1.5)
Lollipops: y=.07(15)
We just input our information into the equations. Using logic, we know that we will have to multiply the cost of the candy by the number of candies to get the total of the three types.

Totals:

Mints: y=.72
Chocolates: y=7.05
Lollipops: y=1.05.
*Recall that y=total cost of candy for each type.

Now, we just simply add the three costs up to get the total sum that the candy will cost:

.72+7.05+1.05=8.82

Therefore, all the candy will cost $8.82.

You might be interested in

Round 3 to 2-5+4+21+1666666-2times 43

yKpoI14uk [10]

I am so sorry if this is wrong, but
2−5+4+21+1666666−(2)(43)
=−3+4+21+1666666−(2)(43)
=1+21+1666666−(2)(43)
=22+1666666−(2)(43)
=1666688−(2)(43)
=1666688−86
Your answer would be 1,666,602
Hope this helps!:)

6 0

3 years ago

A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an a

Scrat [10]

Answer:

The answer is (C) 8

Step-by-step explanation:

First, let's calculate the length of the side of the square.

$A_{square}=a^2$ , where $a$ is the length of the side. Now, let's try to build the square. First we need to find a point which distance from (0, 0) is 10. For this, we can use the distance formula in the plane:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ which for $x_1=0$ and $y_1 = 0$ transforms as $d=\sqrt{(x_2)^2 + (y_2)^2}$ . The first point we are looking for is connected to the origin and therefore, its components will form a right triangle in which, the Pythagoras theorem holds, see the first attached figure. Then, $x_2$ , $y_2$ and 10 are a Pythagorean triple. From this, $x_2= 6$ or $x_2=8$ while $y_2= 6$ or $y_2=8$ . This leads us with the set of coordinates:

$(\pm 6, \pm 8)$ and $(\pm 8, \pm 6)$ . (A)

The next step is to find the coordinates of points that lie on lines which are perpendicular to the lines that joins the origin of the coordinate system with the set of points given in (A):

Let's do this for the point (6, 8).

The equation of the line that join the point (6, 8) with the origin (0, 0) has the equation $y = mx +n$ , however, we only need to find its slope in order to find a perpendicular line to it. Thus,

$m = \frac{y_2-y_1}{x_2-x_1} \\m = \frac{8-0}{6-0} \\m = 8/6$

Then, a perpendicular line has an slope $m_{\bot} = -\frac{1}{m} = -\frac{6}{8}$ (perpendicularity condition of two lines). With the equation of the slope of the perpendicular line and the given point (6, 8), together with the equation of the distance we can form a system of equations to find the coordinates of two points that lie on this perpendicular line.

$m_{\bot}=\frac{6}{8} = \frac{8-y}{6-x}\\ 6(6-x)+8(8-y)=0$ (1)

$d^2 = \sqrt{(y_o-y)^2+(x_o-x)^2} \\(10)^2=\sqrt{(8-y)^2+(6-x)^2}\\100 = \sqrt{(8-y)^2+(6-x)^2}$ (2)

This system has solutions in the coordinates (-2, 14) and (14, 2). Until here, we have three vertices of the square. Let's now find the fourth one in the same way we found the third one using the point (14,2). A line perpendicular to the line that joins the point (6, 8) and (14, 2) has an slope $m = 8/6$ based on the perpendicularity condition. Thus, we can form the system:

$\frac{8}{6} =\frac{2-y}{14-x} \\8(14-x) - 6(2-y) = 0$ (1)

$100 = \sqrt{(14-x)^2+(2-y)^2}$ (2)

with solution the coordinates (8, -6) and (20, 10). If you draw a line joining the coordinates (0, 0), (6, 8), (14, 2) and (8, -6) you will get one of the squares that fulfill the conditions of the problem. By repeating this process with the coordinates in (A), the following squares are found:

(0, 0), (6, 8), (14, 2), (8, -6)

(0, 0), (8, 6), (14, -2), (6, -8)
(0, 0), (-6, 8), (-14, 2), (-8, -6)
(0, 0), (-8, 6), (-14, -2), (-6, -8)

Now, notice that the equation of distance between the two points separated a distance of 10 has the trivial solution $(\pm10, 0)$ and $(0, \pm10)$ . By combining this points we get the following squares:

(0, 0), (10, 0), (10, 10), (0, 10)
(0, 0), (0, 10), (-10, 10), (-10, 0)
(0, 0), (-10, 0), (-10, -10), (0, -10)
(0, 0), (0, -10), (-10, -10), (10, 0)

See the attached second attached figure. Therefore, 8 squares can be drawn

8 0

3 years ago

A water storage tank has the shape of a cylinder with diameter 14 ft. It is mounted so that the circular cross-sections are vert

andrezito [222]

Answer:

percentage of the total capacity is 75.6%

Step-by-step explanation:

Hello! To solve this problem we follow the following steps

1. draw the complete scheme of the problem (see attached image)

2. To solve this problem we must find the area of the circular sector using the following equation.(c in the second attached image)

$A=\frac{R^2}{2} (\alpha -sin\alpha )$

$\alpha =2arccos(\frac{d}{R})$

3. observing the attached images we replace the values in the equations and find the area of the circular sector, remember that you must transform the angle to radians

$\alpha =2arccos(\frac{5}{7})=88.83$