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Sati [7]
3 years ago
9

A gardener buys a plant that is 12cm in height. Each week after that the plant grows 10cm. Note: The plant is 12cm high at the b

eginning of the week.
What will be the height of the plant at the beginning of the 1st, 2nd, 3rd, 4th, and 5th weeks, if it follows the same pattern?

Please show your work. Thank you for your help!!!!
Mathematics
2 answers:
BaLLatris [955]3 years ago
6 0
82, 152, 222, 292, 362.  You add 70 each time.
zhannawk [14.2K]3 years ago
6 0
You have to add 70 to each
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a glass of jar contains 1 red, 3 green and 2 blue marbles. asingle marble is chosen at random from the jar and its color is reco
lubasha [3.4K]

hi,  

first let's count the marbles :  1+3+2 = 6  

so picking a red  is  1/6

                  a green is :  3/6

                    a bleu is :  2/6

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What is the measure of &lt;2?
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I think 90 degrees but not 100 percent sure
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3 years ago
TIME SENSITIVE! PLEASE HELP!
Readme [11.4K]

Answer:

Answer choice A

Step-by-step explanation:

For the first letter, there are 26 options, as there are for the second and third letter. For each of the three numbers, there are 10 digits to choose from. Therefore, 26*26*26*10*10*10=17,576,000, or answer choice A. Hope this helps!

7 0
3 years ago
The arch beneath a bridge is​ semi-elliptical, a​ one-way roadway passes under the arch. The width of the roadway is 38 feet and
forsale [732]

Answer:

Only truck 1 can pass under the bridge.

Step-by-step explanation:

So, first of all, we must do a drawing of what the situation looks like (see attached picture).

Next, we can take the general equation of an ellipse that is centered at the origin, which is the following:

\frac{x^2}{a^2}+\frac{y^2}{b^2}

where:

a= wider side of the ellipse

b= shorter side of the ellipse

in this case:

a=\frac{38}{2}=19ft

and

b=12ft

so we can go ahead and plug this data into the ellipse formula:

\frac{x^2}{(19)^2}+\frac{y^2}{(12)^2}

and we can simplify the equation, so we get:

\frac{x^2}{361}+\frac{y^2}{144}

So, we need to know if either truk will pass under the bridge, so we will match the center of the bridge with the center of each truck and see if the height of the bridge is enough for either to pass.

in order to do this let's solve the equation for y:

\frac{y^{2}}{144}=1-\frac{x^{2}}{361}

y^{2}=144(1-\frac{x^{2}}{361})

we can add everything inside parenthesis so we get:

y^{2}=144(\frac{361-x^{2}}{361})

and take the square root on both sides, so we get:

y=\sqrt{144(\frac{361-x^{2}}{361})}

and we can simplify this so we get:

y=\frac{12}{19}\sqrt{361-x^{2}}

and now we can evaluate this equation for x=4 (half the width of the trucks) so:

y=\frac{12}{19}\sqrt{361-(8)^{2}}

y=11.73ft

this means that for the trucks to pass under the bridge they must have a maximum height of 11.73ft, therefore only truck 1 is able to pass under the bridge since truck 2 is too high.

5 0
2 years ago
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