Which is an algebraic expression for five more then Z?

Mathematics

2 answers:

algol [13]3 years ago

5 0

5 > Z more than is always like this > more than or equal to is like this > with a line under it <span />

faltersainse [42]3 years ago

4 0

I believe the answer would be Z < (Z+5)

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A measuring cylinder of radius 3cm contains water to a height of 49cm. if this water is poured into a similar cylinder of radius

levacccp [35]

Hey there! I'm happy to help!

First, let's find the area of the circle of the first cylinder. To do this, we square the radius and multiply by 3.14.

3²=9

9×3.14=28.26

Now we multiply by our height of 49 cm to see how much water there is.

28.26×49=1384.74

Now, let's find the area of the circle of the 7 cm cylinder.

7²=49

49×3.14=153.86

Now, we will take our liquid amount (volume) and divide by this circle amount (base) to find the number of centimeters of height of the water column.

1384.74÷153.86=9

So, the height of our water column will be 9 cm.

Have a wonderful day and keep on learning! :D

5 0

3 years ago

Use an Addition or Subtraction Formula to find the exact value of the expression, as demonstrated in Example 1. cos 41π 12.

PtichkaEL [24]

Answer: $\frac{\sqrt{6}-\sqrt{2}}{2}$

Step-by-step explanation:

We apply the formula $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$ .

Note that $\cos(\frac{41}{12}\pi)=\cos((\frac{36}{12}+\frac{7}{12})\pi)=\cos(3\pi + \frac{7}{12})\pi)$ . Take $x=3\pi$ and $y=\frac{7}{12}\pi$ in the formula above to get

$\cos(\frac{41}{12}\pi)=\cos(3\pi)\cos(\frac{7}{12}\pi)-\sin(3\pi)\sin(\frac{7}{12}\pi)=(-1)\cdot \cos(\frac{7}{12}\pi)-0\cdot\sin(\frac{7}{12}\pi)=-\cos(\frac{7}{12}\pi)$

Then the value of this expression is $-\cos(\frac{7}{12}\pi)$

We can use the cosine addition formula again to simplify further. Decompose the fraction in the argument as:

$\cos(\frac{7}{12}\pi)=\cos((\frac{3}{12}+\frac{4}{12})\pi)=\cos((\frac{1}{4}\pi + \frac{1}{3})\pi)$

Applying the formula with $x=\frac{1}{4}\pi$ and $y=\frac{1}{3}\pi$ we obtain

$\cos(\frac{7}{12}\pi)=\cos(\frac{1}{4}\pi)\cos(\frac{1}{3}\pi)-\sin(\frac{1}{4}\pi)\sin(\frac{1}{3}\pi)=\frac{\sqrt{2}}{2}\cdot\frac{1}{2} -\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{2}-\sqrt{6}}{2}$