Ask question

Ask question

All categories

2 years ago

14

If I give you seven apples, you will then have five times as many as I would then have, however, if you give me seven apples, we

will then both have the same number of
apples. How many apples do we currently have?

1 answer:

jeka57 [31]2 years ago

3 0

Answer:

28 apples

Step-by-step explanation:

You might be interested in

Devin has a furniture store. He sold a shoe rack recently for 3450. It cost him just 2300 to make the rack

True [87]

Answer:

50%

Step-by-step explanation:

3450 - 2300 = 1150

1150/ 2300 × 100 = 50

5 0

2 years ago

PLEASE HELP im super confused

FinnZ [79.3K]

Answer:

7 Quarters and 3 dimes

Step-by-step explanation:

You have to find how many quarters (0.25) and dimes (0.10) are in $2.05. There are 7 quarters, which will add up to $1.75. You can't have 8 quarters because then you would have 2 dollars, and there wouldn't be enough for a dime, there would only by 5 cents left, or a nickle. Then, you find out how many nickels you need, which is 3.

.10x3=.30. 1.75+0.3=2.05.

I hope you understood:)

4 0

2 years ago

Read 2 more answers

Whats the product of (-3x+2y)(4x-y)

oee [108]

Answer:

-12x^2 + 11xy - 2y^2

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D is the trian

Alla [95]

Answer: mass (m) = 4 kg

center of mass coordinate: (15.75,4.5)

Step-by-step explanation: As a surface, a lamina has 2 dimensions (x,y) and a density function.

The region D is shown in the attachment.

From the image of the triangle, lamina is limited at x-axis: 0≤x≤2

At y-axis, it is limited by the lines formed between (0,0) and (2,1) and (2,1) and (0.3):

<u>Points (0,0) and (2,1):</u>

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

y = $\frac{x}{2}$

<u>Points (2,1) and (0,3):</u>

y = $\frac{3-1}{0-2}(x-0) + 3$

y = -x + 3

Now, find total mass, which is given by the formula:

$m = \int\limits^a_b {\int\limits^a_b {\rho(x,y)} \, dA }$

Calculating for the limits above:

$m = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2(x+y)} \, dy \, dx }$

where a = -x+3

$m = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2} {(xy+\frac{y^{2}}{2} )} \, dx }$

$m = 2.\int\limits^2_0 {(-x^{2}-\frac{x^{2}}{2}+3x )} \, dx }$

$m = 2.\int\limits^2_0 {(\frac{-3x^{2}}{2}+3x)} \, dx }$

$m = 2.(\frac{-3.2^{2}}{2}+3.2-0)$

m = 2(-4+6)

m = 4

<u>Mass of the lamina that occupies region D is 4.</u>

<u />

Center of mass is the point of gravity of an object if it is in an uniform gravitational field. For the lamina, or any other 2 dimensional object, center of mass is calculated by:

$M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }$

$M_{y} = \int\limits^a_b {\int\limits^a_b {x.\rho(x,y)} \, dA }$

$M_{x}$ and $M_{y}$ are moments of the lamina about x-axis and y-axis, respectively.

Calculating moments:

For moment about x-axis:

$M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }$

$M_{x} = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2.y.(x+y)} \, dy\, dx }$

$M_{x} = 2\int\limits^2_0 {\int\limits^a_\frac{x}{2} {y.x+y^{2}} \, dy\, dx }$

$M_{x} = 2\int\limits^2_0 { ({\frac{y^{2}x}{2}+\frac{y^{3}}{3})}\, dx }$

$M_{x} = 2\int\limits^2_0 { ({\frac{x(-x+3)^{2}}{2}+\frac{(-x+3)^{3}}{3} -\frac{x^{3}}{8}-\frac{x^{3}}{24} )}\, dx }$

$M_{x} = 2.(\frac{-9.x^{2}}{4}+9x)$

$M_{x} = 2.(\frac{-9.2^{2}}{4}+9.2)$

$M_{x} = 18$

Now to find the x-coordinate:

x = $\frac{M_{y}}{m}$

x = $\frac{63}{4}$

x = 15.75

For moment about the y-axis:

$M_{y} = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2x.(x+y))} \, dy\,dx }$

$M_{y} = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2} {x^{2}+yx} \, dy\,dx }$

$M_{y} = 2.\int\limits^2_0 {y.x^{2}+x.{\frac{y^{2}}{2} } } \,dx }$

$M_{y} = 2.\int\limits^2_0 {x^{2}.(-x+3)+\frac{x.(-x+3)^{2}}{2} - {\frac{x^{3}}{2}-\frac{x^{3}}{8} } } \,dx }$

$M_{y} = 2.\int\limits^2_0 {\frac{-9x^3}{8}+\frac{9x}{2} } \,dx }$

$M_{y} = 2.({\frac{-9x^4}{32}+9x^{2})$

$M_{y} = 2.({\frac{-9.2^4}{32}+9.2^{2}-0)$

$M{y} =$ 63

To find y-coordinate:

y = $\frac{M_{x}}{m}$

y = $\frac{18}{4}$

y = 4.5

<u>Center mass coordinates for the lamina are (15.75,4.5)</u>

3 0

3 years ago

From Midnight to 6 am the temperature rose 8 degrees C. At 6 am the temperature was -21 degrees C. What was the temperature at m

Alika [10]

Hello!

We know that from midnight to 6am the temperature went up 8° and at 6am it was -21°

To find what it was at midnight we have to work backwards

Since it went up we have to subtract 8 from -21

-21 - 8 = -29

The answer is -29°C

Hope this helps!

7 0

3 years ago

Read 2 more answers