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3 years ago

What is the distance between the points (–3, 4) and (–7, 4)?

Mathematics

2 answers:

Liono4ka [1.6K]3 years ago

6 0

10 I hope this helps bye

Elodia [21]3 years ago

3 0

Distance between (x1,y1) and (x2,y2) is
$D= \sqrt{(x2-x1)^2+(y2-y1)^2}$

given
(-3,4) and (-7,4)

$D= \sqrt{(-7-(-3))^2+(4-4)^2}$
$D= \sqrt{(-7+3)^2+(0)^2}$
$D= \sqrt{(-4)^2+0}$
D=√16
D=4

distance is 4

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Find two integers whose sum is -5 andproduct is 4

nirvana33 [79]

Answer:

-1 and -4

Step-by-step explanation:

-1 + -4 = -5

-1 times -4 = 4

give brainiest please!

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7 0

2 years ago

The school that Scott goes to is selling tickets to a fall musical. On the first day of ticket sales the school sold 12 adult ti

harina [27]

Answer:

The price of one adult ticket = $13

The price of one student ticket = $4

Step-by-step explanation:

Let the price of 1 adult ticket = x

Let the price of 1 student ticket = y

On the first day of ticket sales the school sold 12 adult tickets and 10 student tickets for a total of $196.

12 × x + 10 × y = $196

12x + 10y = 196....... Equation 1

The school took in $59 on the second day by selling 3 adult tickets and 5 student tickets

3 × x + 5 × y = $59

3x + 5y = 59.......... Equation 2

Using Elimination method

We eliminate y, by Multiplying equation 1 by 5 and Equation 2 by 10

12x + 10y = 196....... Equation 1 × 5

3x + 5y = 59.......... Equation 2 × 10

60x + 50y = 980....... Equation 3

30x + 50y = 590......... Equation 4

Subtract Equation 4 from Equation 3

30x = 390

x = 390/30

x = $13

Therefore, the price of one adult ticket = $13

Remember: x = $13

3x + 5y = 59.......... Equation 2

Substitute

3(13) + 5y = 59

39 + 5y = 59

5y = 59 - 39

5y = 20

y = 20/5

y = $4

Therefore, the price of one student ticket = $4

8 0

3 years ago

How do you classify this polynomial? 4x^5-3x^2-8

WINSTONCH [101]

This would be a Quintic Trinomial; it has a degree of 5 and 3 terms.

7 0

3 years ago

What is the value of x in this equation?<br><br> -14 ÷ x = -7

Vilka [71]

Its 2 heres why:
-14/2=-7 since making the 2 a negative would make the answer +7

6 0

4 years ago

I need help with this word problem please

Anika [276]

Answer:

-60 m/s²

Step-by-step explanation:

We are not given the velocity of the plane at landing but we are given velocities at two time periods

V₅, the velocity of the plane 5 seconds after landing is 110 m/s

Plugging this into the velocity equation we get

$\bold{v_5 = a.5 + v_0 = 5a + v_0}$ (1)

Similarly

$\bold{v_{15} = a.15 + v_0 = 15a + v_0}$ (2)

Subtract (1) from $v_{15} - v_5 = (15a + v_0) - (5a + v_0) = (15a-5a) + (v_0-v_0) = 10a$

Therefore $a = \frac{v_{15} - v_5}{10}$

Given $v_{15}= 50 m/s \textrm{ and } v_5 = 110m/s$

this evaluates to

$a = (50-110)/10 =$ -60 m/s²

The negative value is because the plane is actually decelerating which is a negative number

8 0

2 years ago