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3 years ago

To make 6 apples pie, you need 2 pounds of apples. How many pounds of apples do you need to make 10 apple pies?

2 answers:

4 0

For every 6 apple pie, you need 2 pounds of apples.
So, how many pounds of apples do you need for 10 apple pies?
Well, first to figure out this question we first need to make a conversion factor.
I see in the fraction 6/3 we can simplify that to 2/1. Now we got our conversion factor.
We need to invert the fraction in order for the pie unit to cancel out.
1/2 * 10 = 5

You will need 5 pounds of apples to make 10 apple pies.

d1i1m1o1n [39]3 years ago

3 0

You would need 3 pounds and 8 ounces of apples to make 10 pies.

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At 3:30 in the afternoon in mid-September the Kimball Tower casts a shadow about 290 feet long when the sun's rays come down at

ioda

Answer:

203 feet.

Step-by-step explanation:

Please find the attachment.

Let h represent the height of the building.

We have been given that at 3:30 in the afternoon in mid-September the Kimball Tower casts a shadow about 290 feet long when the sun's rays come down at an angle about 35 degrees above the horizontal.

We know that tangent relates opposite side of a right triangle to its adjacent side.

$\text{tan}=\frac{\text{Opposite}}{\text{Adjacent}}$

Upon substituting our given values in above formula, we will get:

$\text{tan}(35^{\circ})=\frac{h}{290}$

$290*\text{tan}(35^{\circ})=\frac{h}{290}*290$

$290*0.70020753821=h$

$h=290*0.70020753821$

$h=203.0601860809$

$h\approx 203$

Therefore, the building is approximately 203 feet high.

3 0

3 years ago

) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle

Fynjy0 [20]

The surface (call it $S$ ) is a triangle with vertices at the points

$x=0,y=0\implies z=2\implies(0,0,2)$

$x=0,y=2\implies z=-2\implies(0,2,-2)$

$x=2,y=0\implies z=-8\implies(2,0,-8)$

Parameterize $S$ by

$\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)$

with $0\le u\le1$ and $0\le v\le1$ . Take the normal vector to $S$ to be

$\vec s_v\times\vec s_u=(20v,8v,4v)$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv$

$=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}$

8 0

3 years ago

A triangle has an angle that measures 90 degrees what type of triangle could it be

vekshin1

A triangle that has an angle that measures 90 degrees would be a Right Triangle.

6 0

3 years ago

Read 2 more answers

Find an equation for the perpendicular bisector of the line segment whose endpoints are (8,-5)(8,−5) and (4,3)(4,3).

Marysya12 [62]

Answer:

$Correct answer\\y=\frac{1}{2} x-4$

Step-by-step explanation:

$y-\frac{-5+3}{2} =\frac{-1}{\frac{3-(-5)}{4-8} } (x-\frac{4+8}{2} )\\y+1=\frac{1}{2} (x-6)\\y=\frac{1}{2} x-4$

6 0

2 years ago

Pls solve part b) iii thanks

Viktor [21]

The bearing of the tree from Q is 296.565°

<h3>How to determine the height of the tree?</h3>

The figure that illustrates the bearing and the distance is added as an attachment

The given parameters are:

Base of the tree, b = 50 meters

Angle (x) = 32 degrees

Calculate the height (h) of the tree using:

tan(x) = height/base

So, we have:

tan(32°) = h/50

Make h the subject

h= 50 × tan(32°)

Evaluate

h = 31.24

Hence, the height of the tree is 31.24 meters

<h3>How to determine the distance between Q and the base of the tree?</h3>

The distance (d) between Q and the base of the tree

This is calculated using the following Pythagoras theorem

d = √(100² + 50²)

Evaluate

d = 111.80

Hence, the distance between Q and the base of the tree is 111.80 meters

<h3>How to determine the angle of elevation?</h3>

The angle of elevation (x) using the following tangent trigonometric ratio

tan(x) = h/d

This gives

tan(x) = 31.24/111.80

Evaluate the quotient

tan(x) = 0.2794

Take the arc tan of both sides

x = 15.61

<h3>The bearing of the tree from Q </h3>

This is calculated using:

Angle of bearing = 270 + arctan(50/100)

Evaluate the arc tan

Angle of bearing = 270 + 26.565

Evaluate the sum

Angle of bearing = 296.565

Hence, the bearing of the tree from Q is 296.565 degrees