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2 years ago

600+600-200 Giving Brainliest

Mathematics

2 answers:

Mila [183]2 years ago

7 0

Equation Answer is 1,000!

PilotLPTM [1.2K]2 years ago

6 0

A good way to go about this is PEMDAS
P- parentheses
E - exponents
M - Multiplication
D- Divide
A - Addition
S - Subtract
since addition is before subtraction you would add 600+600= 1000 than you would subtract! 1000-200 = 800

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choose the equation of a line that is perpendicular to the y axis and passes through the point (5,-2)

kondaur [170]

Answer:

y = -2

Step-by-step explanation:

The line that is perpendicular to y-axis will be a horizontal line so it's slope is 0 => y = 0x + b or y = b

Using (5,-2) => -2 = b => b = -2

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2 years ago

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The answer is 200 because 100x2 is 200

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1 year ago

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Use the figure to answer the questions below.

alina1380 [7]

Answer:

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Step-by-step explanation:

A. Opposite sides are the same length, so the figure is a parallelogram.

B. Diagonals of a parallelogram cross at their midpoints, so ...

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2 years ago

Suppose we want to choose 5 colors, without replacement, from 8 distinct colors.1. If the order is relevant, how many can be don

Charra [1.4K]

(1) From the information given, if we want to choose 5 colors from 8 distinct colors and the order in which the selection is made is relevant, then what we have is a permutation.

The formula is given as;

$nP_r=\frac{n!}{(n-r)!}$

This formula means we need to select/arrange r items out of a total of n items and the anwer derived would be the total number of arrangements possible.

Therefore, we would have;

$\begin{gathered} nP_r\Rightarrow_8P_5 \\ _8P_5=\frac{8!}{(8-5)!}\Rightarrow\frac{8!}{3!} \\ _8P_5=\frac{8\times7\times6\times\ldots1}{3\times2\times1}\Rightarrow\frac{40320}{6} \\ _8P_5=6720 \end{gathered}$

Therefore, if the order is relevant, this selection can be done in 6,720 ways.

(2) If the order is NOT relevant, then what we need to calculate is a combination and the formula is;

$_nC_r=\frac{n!}{(n-r)!r!}$

The formula can now be applied as follows;

$\begin{gathered} _nC_r\Rightarrow_8C_5 \\ _8C_5=\frac{8!}{(8-5)!\times5!} \\ _8C_5=\frac{8!}{3!\times5!}\Rightarrow\frac{8\times7\times6\times\ldots1}{(3\times2\times1)\times(5\times4\times\ldots1)} \\ _8C_5=\frac{40320}{6\times120} \\ _8C_5=56 \end{gathered}$

If the order is not relevant, then the selection can be done in 56 ways.

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tensa zangetsu [6.8K]

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600+600-200 Giving Brainliest​

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