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3 years ago

To increase or decrease an amount by 30% what single multiplier would you use?

Mathematics

2 answers:

adelina 88 [10]3 years ago

8 0

<span>i think that would be x 30/100</span>

Nataly_w [17]3 years ago

5 0

100%-30%=70%
100%+30%=130%

to decrese by 30%, multiply by 70%
to increase by 30%, multiply by 130%

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The sum of two numbers is -6. The difference is 16. Find the two numbers.

Bess [88]

Answer:

$-5\:\mathrm{and}\: 11$

Step-by-step explanation:

We can write the following system of equations:

$\begin{cases}x+y=6\\x-y=16\\\end{cases}$

Adding both equations together, we isolate $x$ and get:

$2x=22,\: x=11$

Plugging $x=11$ in any of the equations, we can solve for $y$ :

$11+y=6,\: y=-5$

Verify that the solution pair $(11, -5)$ works $\checkmark$

Therefore, the two numbers are $\fbox{$-5\:\mathrm{and}\: 11$}$ .

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3 years ago

Using Heron’s formula, calculate the area of the parallelogram to the nearest tenth of a square unit.

11Alexandr11 [23.1K]

Answer:

the area = 36.7

Step-by-step explanation:

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3 years ago

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What the heck do i do with this

Ad libitum [116K]

Use the law of cosines to find a: $a^2=b^2+c^2-2bc (cos(A))$

So to use the given numbers you get: $a^2=27^2+45^2-2(27)(45)(cos(53))$

Which would come out to equal about 35.94

so a=35.9

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3 years ago

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I don't know how to do this (answer pls ) number 3

Blababa [14]

Subtract 2b by both sides which gets -b+4=-5 then subtract 4 by both sides so you get -b=-9 and divide -1 by both sides so that b=9

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3 years ago

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Homework Jim Thompson! PART 2

BlackZzzverrR [31]

Problem 2

Part (a)

The 3D shape formed when rotating around the y axis forms a pencil tip

The shape formed when rotating around the x axis is a truncated cone turned on its side.

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Part (b)

Check out the two diagrams below.

============================================================

Problem 3

Answer: Choice A and Choice C

-----------------------

Explanation:

Think of stacks of coins. Let's say we had 2 stacks of 10 quarters each. The quarters are identical, so they must produce identical volumes. Those sub-volumes then add up to the same volume for each stack. Now imagine one stack is perfectly aligned and the other stack is a bit crooked. Has the volume changed for the crooked stack? No, it hasn't. We're still dealing with the same amount of coins and they yield the same volume.

For more information, check out Cavalieri's Principle.

With all that in mind, this leads us to choice C. If the bases are the same, and so are the heights, then we must be dealing with the same volumes.

On the other hand, if one base is wider (while the heights are still equal) then the wider based block is going to have more volume. This leads us to choice A.

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2 years ago