Ask question

Ask question

All categories

kondor19780726 [428]

3 years ago

11

Find the distance between the points (4,3) and (2,-1). Express your answer in simplest radical form.

1 answer:

Kamila [148]3 years ago

3 0

Answer : B

the distance between the points (4,3) and (2,-1)

We apply distance formula

$d^2= (x_2-x1)^2 + (y_2-y_1)^2$

(x1,y1) is (4,3)

(x2,y2) is (2,1)

$d^2= (2-4)^2 + (-1-3)^2$

$d^2= (-2)^2 + (-4)^2$

d^2 = 4+16 = 20

Take square root on both sides

So $d=2\sqrt{5}$

the distance between the points (4,3) and (2,-1) is 2sqrt(5)

You might be interested in

find the area of field in a form of trapezium whose parallel sides are 10 and 12 and perpendicular distance between them is 6 me

IRISSAK [1]

Answer:

66 m²

Step-by-step explanation:

A = ½ (a + b) h

A = ½ (10 + 12) (6)

A = 66

The area is 66 m².

7 0

3 years ago

Help me plzzzzzzzzz!!!!!!!!!!!!!!!!!!

Anvisha [2.4K]

Answer: ( 12 , 9 )

Step-by-step explanation:

The formula for finding the coordinate of points when a line is externally divided in a given ratio is given by :

x = $\frac{p x_{2}+q x_{1}}{p+q}$

y = $\frac{p y_{2}+q y_{1}}{p+q}$

From the question

$x_{1}$ = 2

$x_{2}$ = ?

x = 4

$y_{1}$ = -1

$y_{2}$ = ?

y = 1

p = 1

q = 4

substituting the values into the formula ,we have

x = $\frac{p x_{2}+q x_{1}}{p+q}$

4 = $\frac{1(x_{2})+4(2) }{1+4}$

4 = $\frac{x_{2}+8}{5}$

$x_{2}$ + 8 = 20

$x_{2}$ = 12

Also

y = $\frac{p y_{2}+q y_{1}}{p+q}$

1 = $\frac{y_{2}+4(-1)}{5}$

$y_{2}$ - 4 = 5

$y_{2}$ = 9

Therefore , the end of the stencil is located at point ( 12 , 9 )

5 0

3 years ago

Graph the line y = kx +1 if it is known that the point M belongs to it: M(1,3)

tigry1 [53]

Answer:

$M(x,y) = (1,3)$ belongs to the line $y = 2\cdot x +1$ . Please see attachment below to know the graph of the line.

Step-by-step explanation:

From Analytical Geometry we know that a line is represented by this formula:

$y=k\cdot x + b$

Where:

$x$ - Independent variable, dimensionless.

$y$ - Dependent variable, dimensionless.

$k$ - Slope, dimensionless.

$b$ - y-Intercept, dimensionless.

If we know that $b = 1$ , $x = 1$ and $y = 3$ , then we clear slope and solve the resulting expression:

$k = \frac{y-b}{x}$

$k = \frac{3-1}{1}$

$k = 2$

Then, we conclude that point $M(x,y) = (1,3)$ belongs to the line $y = 2\cdot x +1$ , whose graph is presented below.

3 0

3 years ago

Select the function that's represented in the graph. Question 5 options: A) ƒ(x) = x2 – 3 B) ƒ(x) = –1∕2x2 – 3 C) ƒ(x) = –x2 – 3

Neporo4naja [7]

Answer:

a)

Step-by-step explanation:

hello,

because of the end behaviour the constant in $x^2$ should be positive so we have a) or d)

f(0)=-3 in both cases

for A) f(x)=

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

so f(x)=0 for $x=\sqrt{3} \ or \ x = -\sqrt{3}$

so the correct answer is A)

hope this helps

3 0

2 years ago

Plz help ma :/<br><br><br> Helpppppppppppppppppppppp

Marizza181 [45]

Answer:

It is 432

Step-by-step explanation:

6 0

3 years ago