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Thepotemich [5.8K]

3 years ago

10

I have a bag of trail mix. One half of the bag is peanuts 1/4 of the bag is chocolate candies,1/4 of the bag is dried fruit. Wha

t is the probability that I will select a peanut?

1 answer:

lana66690 [7]3 years ago

3 0

50% or 1/2 or 1:2 would be your answer

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Sammi bought 2dozen valentines. How many valentines did sammi have left after giving 7/8 of them to her classmates

Ierofanga [76]

Dozen = 12

2 dozen

2x12=24

7/8 of 24

24 / 8 = 3
3 x 7 = 21
24- 21= 3

3 roses are left over

8 0

3 years ago

What is the altitude of the equilateral triangle?

MAXImum [283]

It’s not going to be C because 10 is already in the question

7 0

2 years ago

Which point lies on the graph of Y= 4x-4<br> 1.(0,4)<br> 2.(0,-4)<br> 3.(-4,0)<br> 4.(4,0)

statuscvo [17]

Answer:

<h2>2. (0, -4)</h2>

Step-by-step explanation:

$y=4x-4\\\\\text{Substitute the coordinates of each points to the equation}\\\text{and check the equality.}\\\\1.\ (0,\ 4)\to x=0,\ y=4\\\\4=4(0)-4\\4=0-4\\4=-4\qquad\bold{FALSE}\\\\2.\ (0,\ -4)\to x=0,\ y=-4\\\\-4=4(0)-4\\-4=0-4\\-4=-4\qquad\bold{TRUE}\\\\3.\ (-4,\ 0)\to x=-4,\ y=0\\\\0=4(-4)-4\\0=-16-4\\0=-20\qquad\bold{FALSE}\\\\4.\ (4,\ 0)\to x=4,\ y=0\\\\0=4(4)-4\\0=16-4\\0=12\qquad\bold{FALSE}$

5 0

3 years ago

Help. The problem is in the picture

inn [45]

Answer:

The answer to your question is y = 2/3x + 4

Step-by-step explanation:

Data

Equation y = 2/3 x + 5

New line = ?

Process

Two lines are parallel if the have the same slope. The slope is the coefficient of the x.

1.- Get the slope of the original line

y = 2/3x + 5

slope = 2/3

2.- Two find the equation of a parallel line we need to point I will use (0, 4)

Use the point slope formula

y - y1 = m(x - x1)

y - 4 = 2/3(x - 0)

y - 4 = 2/3 x

y = 2/3x + 4

6 0

3 years ago

Luden [163]

Cone details:

height: h cm

radius: r cm

Sphere details:

radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

<u>Using Pythagoras Theorem</u>

(a)

TO² + TU² = OU²

(h-10)² + r² = 10² [insert values]

r² = 10² - (h-10)² [change sides]

r² = 100 - (h² -20h + 100) [expand]

r² = 100 - h² + 20h -100 [simplify]

r² = 20h - h² [shown]

r = √20h - h² ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (\sqrt{20h - h^2})^2 \ ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20h - h^2) (h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20 - h) (h) ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)$

To find maximum/minimum, we have to find first derivative.

(c)

<u>First derivative</u>

$\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )$

<u>apply chain rule</u>

$\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}$

<u>Equate the first derivative to zero, that is V'(x) = 0</u>

$\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0$

$\Longrightarrow \sf 40h-3h^2=0$

$\Longrightarrow \sf h(40-3h)=0$

$\Longrightarrow \sf h=0, \ 40-3h=0$

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$ <u />

<u>maximum volume:</u> <u>when h = 40/3</u>

$\sf \Longrightarrow max= \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )$

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

<u>minimum volume:</u> <u>when h = 0</u>

$\sf \Longrightarrow min= \dfrac{1}{3} \pi (0)^2(20-0)$

$\sf \Longrightarrow minimum=0 \ cm^3$

6 0

2 years ago

Read 2 more answers