Answer: 1 29/51
Step-by-step explanation:
Answer: rectangle does not have all equal sides
Step-by-step explanation:
the square and rhombus are quadrilaterals that have equal side lengths on all sides
Not an expertise on infinite sums but the most straightforward explanation is that infinity isn't a number.
Let's see if there are anything we missed:
∞
Σ 2^n=1+2+4+8+16+...
n=0
We multiply (2-1) on both sides:
∞
(2-1) Σ 2^n=(2-1)1+2+4+8+16+...
n=0
And we expand;
∞
Σ 2^n=(2+4+8+16+32+...)-(1+2+4+8+16+...)
n=0
But now, imagine that the expression 1+2+4+8+16+... have the last term of 2^n, where n is infinity, then the expression of 2+4+8+16+32+... must have the last term of 2(2^n), then if we cancel out the term, we are still missing one more term to write:
∞
Σ 2^n=-1+2(2^n)
n=0
If n is infinity, then 2^n must also be infinity. So technically, this goes back to infinity.
Although we set a finite term for both expressions, the further we list the terms, they will sooner or later approach infinity.
Yep, this shows how weird the infinity sign is.
Answer:
The density is equal to the weight divided by the volume:
![\rho=\dfrac{W}{V}](https://tex.z-dn.net/?f=%5Crho%3D%5Cdfrac%7BW%7D%7BV%7D)
The paperweight weights 300 grams.
Step-by-step explanation:
The density of the paperweight can be calculated knowing the weight and the volume of this paperweight.
The density is equal to the weight divided by the volume:
![\rho=\dfrac{W}{V}](https://tex.z-dn.net/?f=%5Crho%3D%5Cdfrac%7BW%7D%7BV%7D)
We know that the density of this paperweight is 1.5 grams per cm3, and its volume is 200 cm3.
We can use the formula of the density to calculate the weight:
![\rho=\dfrac{W}{V}\\\\\\W=\rho\cdot V=1.5\;\dfrac{g}{cm^3}\,\cdot\;200\; cm^3\\\\\\W=300\;g](https://tex.z-dn.net/?f=%5Crho%3D%5Cdfrac%7BW%7D%7BV%7D%5C%5C%5C%5C%5C%5CW%3D%5Crho%5Ccdot%20V%3D1.5%5C%3B%5Cdfrac%7Bg%7D%7Bcm%5E3%7D%5C%2C%5Ccdot%5C%3B200%5C%3B%20cm%5E3%5C%5C%5C%5C%5C%5CW%3D300%5C%3Bg)
The paperweight weights 300 grams.