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Ipatiy [6.2K]
3 years ago
8

Determine which expressions could be used to help find each component of the ice cream cone. ( answer choices are in picture que

stion 2 )

Mathematics
1 answer:
grigory [225]3 years ago
3 0

Answer:

1). a

2). b

3). c

4). d

Step-by-step explanation:

From the picture of a cone attached,

Radius of the circular top = r

Height of the cone = h

Slant height of the cone 'l' = 6

Area of the circular top of the cone with radius 'r' = πr²

Amount of space inside the cone = Volume of the cone = \frac{1}{2}\pi r^{2} h

The amount of wrapper needed to cover the side of the ice cream cone, not including the circular top of the cone = πrl = πr(6 in.)

The vertical height 'h' of the cone when the cone is standing upright = \sqrt{6^2-r^2}

[By Pythagoras theorem, l² = h² + r² ⇒ h = \sqrt{l^2-r^2}]

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Y-4=-5 (x+3) in point slope form
aniked [119]
The answer is C, slope = -5; point = (-3, 4). 

The point slope form of a line is y - y₁ = m (x - x₁), with (x₁, y₁) being the coordinates of one of the points of a line. Here, the first step is to find out what the slope is. The slope is the value in front of (x + 3), which is -5. Now we need to find the coordinates of the point. Since -y₁ is equal to -4, the coordinate of the y value is 4. Now onto the x value. Since -x₁ = 3, the coordinate of the x value is -3. That means that the answer is C, slope = -5; point = (-3, 4). 
7 0
3 years ago
99 POINTS PLZ HELP-Different shapes are drawn on cards and then the cards are placed in a bag. The number of cards for each shap
larisa86 [58]

total number of cards: 12 + 27 + 91 = 130

total number of cards with a circle: 91

 so you have a 91/130 probability which reduces to 7/10 probability

3 0
3 years ago
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Based on a research​ survey, when 1020 adults were asked about hand​ hygiene, 44% said that they wash their hands after using pu
aleksley [76]

Considering that the subjects are chosen without replacement, they are not independent, and the probability cannot be found using the binomial distribution.

The binomial distribution and the hypergeometric distribution are quite similar, as:

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  • The difference is that the binomial distribution is for independent trials, that is, in each trial, the probability of success is the same, while the hypergeometric distribution is for dependent trials.
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A similar problem is given at brainly.com/question/21772486

7 0
2 years ago
In a woodworking class, you need to make five triangles out of plywood. Each triangle has a height of 6 inches and a base of 14
vampirchik [111]
The area of a triangle can be calculated through the equation,
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                            A = (14 in)(6 in) / 2 = 42 in²
Since, there are five of these triangles, the total area is,
                            total area = 5(42 in²) = 210 in²
Thus, the total area of the plywood needed is 210 in².
6 0
3 years ago
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Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess
Zinaida [17]

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill

\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2

4 0
3 years ago
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