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3 years ago

What is the length of the hypotenuse of the triangle?

Mathematics

2 answers:

Inessa [10]3 years ago

6 0

pythagorean theorem

for a right triangle with legs (2 sides next to the the right angle) length a and b, and hyptonuse (longest side oposite right angle) is c

$a^2+b^2=c^2$

we can see that the legs are 3cm and 7cm

so a=3 and b=7 (you can do a=7 and b=3 but it doesn't matter)

$a^2+b^2=c^2$

$3^2+7^2=c^2$

$9+49=c^2$

$58=c^2$

sqrare root both sides

$\sqrt{58}=c$

$c=\sqrt{58}$

answer is 4th option

Jobisdone [24]3 years ago

3 0

Essentially, we just use the pythagorean theorem to solve:

7^2 + 3^2 = c^2

49 + 9 = c^2

58 = c^2

c = sqrt 58

The answer is Option D.

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Evaluate (a + b)2 for a = 2 and b = 3

IgorLugansk [536]

The answer is (2+3)=5

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The school is selling potted plants as a fundraiser.Kara sold 12 ferns and 8 ivy plants for $260.Paul sold 15 ivy plants and 6 f

vovangra [49]

x=ferns ; y=ivy

12x+8y=260

6x+15y=240

x=60 -(15/6)y

12(60+ (15/6)y + 8y = 260

x=60 -(15/6)y

720 - 30y + 8y = 260

x=60- (15/6)y

-30y +8y= 260-720 ==> -22y= -460

x=60 -(15/6)y

y= (-460)/(-22)= 20.9090909091

y=20.90

x=60 -(15/6)×20.90= 7.75

×(ferns)=7.75$

y(ivy)=20.90$

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3 years ago

Find the length of the missing sides. Round your answers to the nearest tenth. 8 y x 21

wlad13 [49]

Answer:

x = 20.8

y = 22.3

Step-by-step explanation:

tan(21) = 8/x

or, x = 8/tan(21)

or, x = 20.8 (rounded to the nearest tenth)

sin(21) = 8/y

or, y = 8/sin(21)

or, y = 22.3 (rounded to the nearest tenth)

Answered by GAUTHMATH

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3 years ago

Will give crown help

jasenka [17]

Answer:

D is a right angle

B is an accute angle

A and C are obtuse

4 0

3 years ago

A Ferris wheel is 20 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o'clock position

pantera1 [17]

Answer:

233.48s

3.84 min

Step-by-step explanation:

In order to solve this problem, we can start by drawing what the situation looks like. See attached picture.

We can model this situation by making use of a trigonometric function. Trigonometric functions have the following shape:

$y=A cos(\omega t+\phi)+C$

where:

A= amplitude =-20m because the model starts at the lowest point of the trajectory.

f= the function to use, in this case we'll use cos, since it starts at the lowest point of the trajectory.

t= time

$\omega=$ angular speed.

in this case:

$\omega=\frac{2\pi}{T}$

where T is the period, in this case 6 min or

$6min(\frac{60s}{1min})=360s$

so:

$\omega=\frac{2\pi}{360}$

$\omega = \frac{\pi}{160}$

and

$\phi$ = phase angle

C= vertical shift

in this case our vertical shift will be:

2m+20m=22m

in this case the phase angle is 0 because we are starting at the lowest point of the trajectory. So the equation for the ferris wheel will be:

$y=-20 cos(\frac{\pi}{180}t)+22$

Once we got this equation, we can figure out on what times the passenger will be higher than 13 m, so we build the following inequality:

$-20 cos(\frac{\pi}{180}t)+22>13$

so we can solve this inequality, we can start by turning it into an equation we can solve for t:

$-20 cos(\frac{\pi}{180}t)+22=13$

and solve it:

$-20 cos(\frac{\pi}{180}t)=13-22$

$-20 cos(\frac{\pi}{180}t)=-9$

$cos(\frac{\pi}{180}t)=\frac{9}{20}$

and we can take the inverse of cos to get:

$\frac{\pi}{180}t=cos^{-1}(\frac{9}{20})$

which yields two possible answers: (see attached picture)

$\frac{\pi}{180}t=1.104$ or $\frac{\pi}{180}t=5.179$

so we can solve the two equations. Let's start with the first one:

$\frac{\pi}{180}t=1.104$

$t =1.104(\frac{180}{\pi})$

t=63.25s

and the second one:

$\frac{\pi}{180}t=5.179$

$t=5.179(\frac{180}{\pi})$

t=296.73s

so now we can build our possible intervals we can use to test the inequality:

[0, 63.25] for a test value of 1

[63.25,296.73] for a test value of 70

[296.73, 360] for a test value of 300

let's test the first interval:

[0, 63.25] for a test value of 1

$-20 cos(\frac{\pi}{180}(1))+22>13$

2>13 this is false

let's now test the second interval:

[63.25,296.73] for a test value of 70

$-20 cos(\frac{\pi}{180}(70))+22>13$

15.16>13 this is true

and finally the third interval:

[296.73, 360] for a test value of 300

$-20 cos(\frac{\pi}{180}(300))+22>13$

12>13 this is false.

We only got one true outcome which belonged to the second interval:

[63.25,296.73]

so the total time spent above a height of 13m will be:

196.73-63.25=233.48s

which is the same as:

$233.48(\frac{1min}{60s})=3.84 min$

see attached picture for the graph of the situation. The shaded region represents the region where the passenger will be higher than 13 m.

3 0

3 years ago