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3 years ago

What is the absolute value of -645

Mathematics

2 answers:

Liula [17]3 years ago

5 0

-645 you can't simplify any further

Eva8 [605]3 years ago

4 0

645. The absolute value will always be positive.

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3 years ago

Find the distance between the points (0,2)and (-5,-9)The answer must be a whole number or a fully simplified radical expression

LiRa [457]

Answer:

d = $\sqrt{146}$

Step-by-step explanation:

$d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d= \sqrt{(-5-0)^2+(-9-2)^2}\\d= \sqrt{(-5)^2+(-11)^2}\\d= \sqrt{25+121}\\d= \sqrt{146}$

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3 years ago

Please help im so lost Ty btw

klasskru [66]

Answer:

1) gradient (00) (-2 4) = y2-y1 / 2-1 = 4/-2 = -2 m = -2/1 means = m = -2 (negative slope) 2) gradient y2-y1 / x2-x1 = 3-0 / 2-0 = 3/2 = (1 1/2)/1 m = 1 1/2 (positive slope) we use the formula y-values divided by the change in the x-values. The equation of the gradient each goes like this 1) y = -2x as y is at origin nothing else to add The equation of the gradient each goes like this 2) y = 1 1/2x The equation of the point formula 1) we take the y -y1 = m (x +x 1) = y-0 = -2x (x +0) (as m = -2) y = -2(x +0) and The equation of the point formula 2) y - 0 = m ( x +x1) y - 0 = 1 1/2( x +0) = y = 1 1/2( x +0)

5 0

3 years ago

What is the number of subsets of S= {1, 2, 3…10} that contain five element include 3 or 4 but not both?

Ivenika [448]

Answer:

140

Step-by-step explanation:

To construct a subset of S with said property, we have two choices, include 3 in the subset or include four in the subset. These events are mutually exclusive because 3 and 4 can not both be elements of the subset.

First, let's count the number of subsets that contain the element 3.

Any of such subsets has five elements, but since 3 is already an element, we only have to select four elements to complete it. The four elements must be different from 3 and 4 (3 cannot be selected twice and the condition does not allow to select 4), so there are eight elements to select from. The number of ways of doing this is ${}_8C_4=70$ .

Now, let's count the number of subsets that contain the element 4.

4 is already an element thus we have to select other four elements . The four elements must be different from 3 and 4 (4 cannot be selected twice and the condition does not allow to select 3), so there are eight elements to select from, so this can be done in ${}_8C_4=70$ ways.

We conclude that there are 70+70=140 required subsets of S.

7 0

3 years ago