Write six and twenty-three thousandths in standard form

What is the distance between the points (1, 25) and (4, 21)

Elena-2011 [213]

Hello!

To find the distance between the points (1, 25) and (4, 21), you need to use the distance formula.

The distance formula is: $d =\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ . With the distance formula, you have to assign the ordered pairs before substituting them into the formula, which is also used to to find the slope of functions.

$d =\sqrt{(21-25)^{2}+(4-1)^{2}}$

$d =\sqrt{(-4)^{2}+(3)^{2}}$

$d=\sqrt{25}$

$d=5$

Therefore, the distance between the two points is 5 units.

4 0

3 years ago

Math help..........,

photoshop1234 [79]

The answer is option B. x=18

Hope I could help!

4 0

3 years ago

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Point C is the midpoint of AB.point A has coordinates (2,4)and point C has coordinates (5,0) what are the coordinates of point B

devlian [24]

Answer:

The coordinates of point B is (8,-4)

Step-by-step explanation:

Midpoint of a line segment: It is the point on the line segment that divides the segment in two congruent segments.

Since it is given that C is the midpoint of Line segment AB.

Mid-point of a segment of end points $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$ is, C= $(\frac{x_{1}+x_{2}}{2} , \frac{y_{1}+y_{2}}{2})$ .

As, given the coordinates of A and C are (2,4) and (5,0)

then, we have $x_{1}=2$ , $y_{1}=4$ ; $\frac{x_{1}+x_{2}}{2}=5$ and $\frac{y_{1}+y_{2}}{2}=0$

to find the coordinates of $B(x_{2},y_{2})$ .

$\frac{2+x_{2}}{2}=5$

Simplify:

$2+x_{2}=10$

⇒ $x_{2}=8$

to solve for $y_{2}$ we have, $\frac{y_{1}+y_{2}}{2}=0$

or $y_{1}+y_{2}=0$

or $y_{2}= -y_{1}$ = -4

so, the values of $x_{2}=8$ and $y_{2}=-4$

Therefore, the coordinates of point B is (8,-4)

3 0

3 years ago

Among 500 freshmen pursuing a business degree at a university, 323 are enrolled in an economics course, 205 are enrolled in a ma

Aleksandr-060686 [28]

<h3>The probability of selecting a random student who is enrolled in both the courses is 0.280.</h3>

Step-by-step explanation:

Here, the total number of freshman in the university = 500

The number of students enrolled in Economics = n(E) = 323

The number of students enrolled in Mathematics = n(M) = 205

The number of students enrolled in Both Economics and math

= n(E∩M ) = 140

Let F : Event of selecting a student who is enrolled in both the courses

So, from the given data:

$P(F) = \frac{\textrm{Total number of favorable outcomes}}{\textrm{Total number of outcomes}} \\= \frac{\textrm{Total number of students enrolled in BOTH courses}}{\textrm{Total number of students}} \\ = \frac{140}{500} = \frac{7}{25}$

So, the probability of selecting a random students who is enrolled in both the courses is $(\frac{7}{25} ) = 0.280$

4 0

3 years ago

Amy wants to buy a pair of jeans that cost $65. Today the store is offering a straight discount of 25%. Calculate her savings an

likoan [24]

it should be $81 bucks ..your welcome

6 0

3 years ago

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