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3 years ago

What is the value of x? 2 units 3 units 5 units 8 units

Mathematics

1 answer:

irinina [24]3 years ago

5 0

Answer:B

Step-by-step explanation:

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Write an equation perpendicular to the line 6x+12y=24 and passing through (4,0)

noname [10]

Y = 2x - 8 is perpendicular to 6x+12y=24 and passes through (4,0)

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3 years ago

A printer is printing photos. For every 6 photos, the printer takes 3 minutes.

ki77a [65]

Answer:

Number of Photos - 6, 8, 12, 14, 18

Time (Minutes) - 3, 4, 6, 8, 9

hope this helps.

4 0

2 years ago

CAN ANYONE HELP ME SOLVE THIS?

777dan777 [17]

Step-by-step explanation:

m/1 = ( 180 - 65 - 46 ) - 180

= 180 - 59 = 121

m/2 = 180 - 121 = 59

m/3 = ( 180 - 142 ) - 180

= 180 - 32 - 82

= 66

m/4 = 66

m/5 = 180 - m/2 - m/4

= 180 - 59 - 66

= 180 - 125 = 55

The sum of interior angles of triangles = 180°

8 0

2 years ago

When Ricardo was 9 years old, he was 56 inches tall. Now he is 12 years old, and

Vesna [10]

your answer should be 10.7%

62-56 = 62. 6/563. 0.1071 = 10.7%

8 0

2 years ago

A pebble is dropped into a deep well and 3s later the sound of a splash is heard as the pebble reaches the bottom of the well. T

ankoles [38]

Answer:

$2.89\ \text{s}$

0.11 seconds

$37.4\ \text{m}$

Step-by-step explanation:

$t_1$ = Time taken by the pebble to hit the water

$t_2$ = Time taken by the sound to travel to the observer

$t_1+t_2$ = Time taken by the observer to listen to the splash after stone is thrown = 3 s

$t_1=3-t_2$

a = g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$u_s$ = Speed of sound = 340 m/s

The distance traveled by the pebble

$s=ut+\dfrac{1}{2}at_1^2\\\Rightarrow s=0+\dfrac{1}{2}\times9.81(3-t_2)^2\\\Rightarrow s=4.905(9+t_2^2-6t_2^2)$

Distance traveled by the sound

$s=u_st\\\Rightarrow s=340\times t_2$

The distance traveled by the sound and the pebble is equal

$4.905(9+t_2^2-6t_2)=340\times t_2\\\Rightarrow 9+t_2^2-6t_2=\dfrac{340}{4.905}t_2\\\Rightarrow 9+t_2^2-6t_2=69.32t_2\\\Rightarrow t_2^2-75.32t_2+9=0\\\Rightarrow t_2=\frac{-\left(-75.32\right)\pm \sqrt{\left(-75.32\right)^2-4\times \:1\times \:9}}{2\times \:1}\\\Rightarrow t_2=75.2,0.11$

Since $t_1+t_2=3$ the value of $t_2$ cannot be greater than 3.

So, time taken by the sound to reach the observer is 0.11 seconds.

Time taken by the pebble to hit the water is $t_1=3-t_2=3-0.11=2.89\ \text{s}$

Depth of the well is $s=340t_2=340\times 0.11=37.4\ \text{m}$

8 0

2 years ago