Ask question

Ask question

All categories

3 years ago

8

Jackson recorded the growth of a plant over 10 weeks. The equation y=0.25+4 represents the height y in inches over time x in wee

ks. How tall will the plant be in 12 weeks.

1 answer:

Dmitriy789 [7]3 years ago

6 0

Answer: the plant will 7 inches tall by week 12

Step-by-step explanation: 0.25 multiplied by 12 is 3 then you add the 4.

You might be interested in

A tree was 17 feet tall when it was planted. It grew 8 times that height in 15 years.

Nataliya [291]

Answer:

A. 119 feet

Step-by-step explanation:

the height of tree right now =

17 × 8 = 136 feet

the differences = 136 -17 = 119 feet

6 0

3 years ago

Read 2 more answers

Can you help me with this question please

vaieri [72.5K]

To get the answer, divide 55 by 610. the answer would be 11.09 i’m pretty sure

5 0

3 years ago

samuel has read 3/6 of his assignment.judy has read 4/8 0f her assingment. their assignments were the same size.which sentence i

ladessa [460]

I'm taking you at your word. I think these sentences are true
because you said them:

-- Samuel has read 3/6 of his assignment.

-- Judy has read 4/8 0f her assingment.

-- Their assignments were the same size.

8 0

3 years ago

Will give brainliest <br><br> What is the value of x?<br> x = [?]

vichka [17]

Answer:

Step-by-step explanation:

28

3 0

3 years ago

At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is

Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = $\frac{dx}{dt}$

ship B is sailing north at 20 km/h = $\frac{dy}{dt}$

here x and y are the sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find $\frac{dz}{dt}$ at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

$z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we | | get \\$

$z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h$

derivative first equation w . r. to t we get

$2z\frac{dz}{dt} =$ $-2(130-x)\frac{dx}{dt}$ $+2y\frac{dy}{dt}$

$\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]$

$\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }$

$= \frac{1100}{85.44}\\ = 12.87km/h$

8 0

3 years ago