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dexar [7]
3 years ago
9

Which table could be a partial set of values for a linear function?

Mathematics
2 answers:
gizmo_the_mogwai [7]3 years ago
8 0
My best bet would be A but i'm not 100 Percent sure, Hope it helps :)
Alexandra [31]3 years ago
8 0

Answer: A

i just took the test and got it right trust me

please mark brainliest answer

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Work your brain this evening y'all!! I had $3.00. My Mom gave me $10.00. My Dad Gave me $30.00. My Aunt & Uncle gave me $100
liraira [26]

Answer:

150 i think

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What is the selling price of a $12,543 boat if a $758 discount is given?
Lelu [443]
Given:
Selling Price: 12,543
Discount: 758

We simply deduct the discount from the original price to get the discounted selling price.

12,543 - 758 = 11,785

The new selling price of the boat is $11,785.

Discount of $758 is 6% of the Original selling price.
758 / 12,543 = 0.06
0.06 x 100% = 6%

8 0
3 years ago
, write an expression should be used to determine the total number of points Amanda scored in Tuesday's game?
jekas [21]

tifhsjruf7sh3bfhchwbe

6 0
3 years ago
2: A bag contains 5 black counters, 4 blue counters, and
drek231 [11]

Answer:

The probability that the counter  was blue is \mathbf{\frac{2}{5}}

Step-by-step explanation:

Number of black Counters = 5

Number of blue Counters = 4

Number of white Counters = 1

We need to write down the probability that the counter  was blue.​

First find Total Counters

Total Counters = Number of black Counters + Number of blue Counters + Number of white Counters

Total Counters = 5+4+1

Total Counters = 10

Now, we need to find probability that the counter taken was blue

The formula used is:

Probability= \frac{Number\:of\:favourable\:outcomes}{Total\:outcomes}

There are 4 blue counters in the back, so Favourable outcomes = 4

Probability= \frac{Number\:of\:favourable\:outcomes}{Total\:outcomes}\\Probability= \frac{4}{10}\\Probability= \frac{2}{5}

The probability that the counter  was blue is \mathbf{\frac{2}{5}}

5 0
3 years ago
A triangle ABC has its vertices at A(-2, -3), B(2, 1), and C(5,-1).
maksim [4K]

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill ~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-3})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{1}) ~\hfill AB=\sqrt{( 2- (-2))^2 + ( 1- (-3))^2} \\\\\\ AB=\sqrt{(2+2)^2+(1+3)^2}\implies AB=\sqrt{32}\implies \boxed{AB=4\sqrt{2}} \\\\[-0.35em] ~\dotfill

B(\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad C(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) ~\hfill BC=\sqrt{( 5- 2)^2 + ( -1- 1)^2} \\\\\\ BC=\sqrt{3^2+(-2)^2}\implies BC=\sqrt{9+4}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad A(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-3}) ~\hfill CA=\sqrt{( -2- 5)^2 + ( -3- (-1))^2} \\\\\\ CA=\sqrt{(-7)^2+(-3+1)^2}\implies CA=\sqrt{49+(-2)^2}\implies \boxed{CA=\sqrt{53}}

\stackrel{\textit{\large perimeter of ABC}}{4\sqrt{2}+\sqrt{13}+\sqrt{53}~~\approx~~ 16.54}

3 0
2 years ago
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