All categories

3 years ago

Which herbal remedy would most likely be used to treat a sunburn?

Mathematics

1 answer:

tangare [24]3 years ago

6 0

You are correct aloe vera is the answer as it has cooling properties <span />

You might be interested in

Which one does not belong and why

Sliva [168]

Answer:

The Circle on the Top right doesn't belong since every other circle has a green quadrant, whereas on the other hand the circle on top right is missing the green quadrant

Hope this helps!

7 0

3 years ago

What are 3 equations equivalent to -19x + 12

lubasha [3.4K]

There is no equation. There is just a polynomial shown

6 0

3 years ago

What is the effect on the graph of the function f(x) = 2x when f(x) is replaced with f(0.5x)? A) vertical stretching B) vertical

PtichkaEL [24]

<h3><u>Answer:</u></h3>

Option: D

Horizontal stretching.

<h3><u>Step-by-step explanation:</u></h3>

We have to find the effect on the graph of the function f(x)=2x when it is replaced by f(0.5 x).

We know that when a parent function f(x) is replaced by f(kx) then either the graph is stretched horizontally or shrinked horizontally.

if k>1 then the graph is shrinked horizontally.

if k<1 then the graph is stretched horizontally.

Hence here k=0.5<1 so the graph of the function is stretched horizontally.

3 0

4 years ago

Read 2 more answers

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$

$=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}$

What we use?

We use that

$e^{i\pi n}=cos(\pi n)+i sin(\pi n)$

and

$\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}$

6 0

4 years ago

Candy's Sweets Company charges $1.15 per pound to ship candy.

allochka39001 [22]

Answer:

Part B:

Step-by-step explanation:

<u>Linear Models</u>

Candy's Sweets Company charges $1.15 per pound to ship candy. This represents a proportional relationship between the pounds of candy and the cost.

Part A: If each pound costs $1.15, then p pounds cost $1.15p. Then the equation of the cost c is:

c = 1.15p

The cost of shipping p=2 pounds of candy is:

c = 1.15*2 = 2.30

c = $2.30

Part B: When the company reduces the cost by $0.25 per pound, the new unit cost is $1.15 - $0.25 = $0.90 per pound.

The new equation to determine the total cost for p pounds of candy is:

c = 0.90p

8 0

3 years ago