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3 years ago

A rocket was launched and its height, h, in metres, above the ground after time,

Mathematics

1 answer:

Marianna [84]3 years ago

6 0

Short answer b 7m

Very simple just do the math mostly multiplying

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Please help! I have no idea what to do. I couldn't even guess an answer.

yulyashka [42]

y - y1 = m(x - x1)

Plug in:

Slope = m = 4

Point = (x1,y1) ----> (-3,7)

y - 7 = 4(x - -3) //Solve for y

y - 7 = 4(x + 3) //Answer

y - 7 = 4x + 12

y = 4x + 12 + 7

y = 4x + 19

Answer: B

//Hope it helps

4 0

3 years ago

The ratio 1 cm and 1.5 km

ehidna [41]

Answer:

1/150000

Step-by-step explanation:

from km to cm you should multiply it by 100000

8 0

3 years ago

Jordan has 5/6 dvds as bob, so if jordan has 55 dvds how much does bob have?

Luba_88 [7]

Answer:

Bob has 66 dvds

Step-by-step explanation:

Let's assume

Bob has 'x' number of dvds

we are given

Jordan has 5/6 dvds as bob

so, Jordon will have

$=\frac{5}{6}x$

now, we are given jordan has 55 dvds

so, we can set it to 55

and then we can solve for x

$\frac{5}{6}x=55$

Multiply both sides by 6

$6\times \frac{5}{6}x=6\times 55$

$5x=6\times 55$

divide both sides by 5

$\frac{5x}{5} =\frac{6\times 55}{5}$

$x=66$

So,

Bob has 66 dvds

3 0

3 years ago

Solve. 1/4 x < -8 PLEASE

Amiraneli [1.4K]

Answer:

x<-32

Step-by-step explanation:

5 0

4 years ago

Read 2 more answers

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

3 years ago