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3 years ago

Tricia made a circular mat from a square piece of cloth as shown in the diagram below.

Mathematics

1 answer:

kap26 [50]3 years ago

8 0

The option C) 49 inches is the circumference of the circular mat.

Step-by-step explanation:

Tricia made a circular mat from a square piece of cloth.
The measure of the side length of the square = 14 inches.
The length of the side is same as the diameter of the circle.
Therefore, the diameter of the circular mat = 14 inches.

To find radius :

Radius = diameter / 2

⇒ 14/2

⇒ 7 inches.

To find the circumference of the circular mat :

Circumference = 2πr

where,

π has the default value of 3.14

r is the radius of the circular mat.

⇒ 2 × 3.14 × 7

⇒ 44 inches.

∴ The circumference of the circular mat is 44 inches.

Since, it is not given in the option. The option C) 49 inches which is close to 44 inches is the right option.

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Please help!! fully simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B300%7D%2B%5Csqrt%7B27%7D-%5Csqrt%7B

bija089 [108]

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3 years ago

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NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h ( t ) = − 4.

Oliga [24]

Answer:

$\displaystyle 1)48.2 \: \: \text{sec}$

$\rm \displaystyle 2)3021.6 \: m$

Step-by-step explanation:

<h3>Question-1:</h3>

so when flash down occurs the rocket will be in the ground in other words the elevation(height) from ground level will be 0 therefore,

to figure out the time of flash down we can set h(t) to 0 by doing so we obtain:

$\displaystyle - 4.9 {t}^{2} + 229t + 346 = 0$

to solve the equation can consider the quadratic formula given by

$\displaystyle x = \frac{ - b \pm \sqrt{ {b}^{2} - 4 ac} }{2a}$

so let our a,b and c be -4.9,229 and 346 Thus substitute:

$\rm\displaystyle t = \frac{ - (229) \pm \sqrt{ {229}^{2} - 4.( - 4.9)(346)} }{2.( - 4.9)}$

remove parentheses:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ {229}^{2} - 4.( - 4.9)(346)} }{2.( - 4.9)}$

simplify square:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 52441- 4( - 4.9)(346)} }{2.( - 4.9)}$

simplify multiplication:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 52441- 6781.6} }{ - 9.8}$

simplify Substraction:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 45659.4} }{ - 9.8}$

by simplifying we acquire:

$\displaystyle t = 48.2 \: \: \: \text{and} \quad - 1.5$

since time can't be negative

$\displaystyle t = 48.2$

hence,

at 48.2 seconds splashdown occurs

<h3>Question-2:</h3>

to figure out the maximum height we have to figure out the maximum Time first in that case the following formula can be considered

$\displaystyle x _{ \text{max}} = \frac{ - b}{2a}$

let a and b be -4.9 and 229 respectively thus substitute:

$\displaystyle t _{ \text{max}} = \frac{ - 229}{2( - 4.9)}$

simplify which yields:

$\displaystyle t _{ \text{max}} = 23.4$

now plug in the maximum t to the function:

$\rm \displaystyle h(23.4)- 4.9 {(23.4)}^{2} + 229(23.4)+ 346$

simplify:

$\rm \displaystyle h(23.4) = 3021.6$

hence,

about 3021.6 meters high above sea-level the rocket gets at its peak?

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