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ki77a [65]
3 years ago
5

Which real-world scenario involves a right triangle?

Mathematics
2 answers:
Likurg_2 [28]3 years ago
7 0

Answer:

I agree with the previous answer that the correct answer is a triangular bike path with lengths of 5 miles, 12 miles, and 13 miles.

Natali [406]3 years ago
5 0

Answer:

A triangular bike path with lengths of 5 miles, 12 miles, and 13 miles

Step-by-step explanation:

The 5-12-13 right triangle is a common pythagorean triple. :)

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The coefficient of 6x is<br> 1<br> 6<br> Х
qwelly [4]

Answer:

6

Step-by-step explanation:

If a number and a variable were together in a term, the number would the the coefficient. The coefficient would multiply the variable.

In '6x', the number '6' is the coefficient. '6' would be multiplying 'x'.

The correct answer should be 6.  

6 0
4 years ago
Read 2 more answers
at an animal shelter there are 15 dogs 12 cats 3 snakes and 5 parakeets. what percent of the number of cats is the number of sna
dsp73
Percentage of cats
\frac{# animals}{#cats} = \frac{100}{?}
\frac{35}{12}=  \frac{100}{?} &#10;
**cross multiply 12 and 100 which is 1200
**divide 35 into 1200
the answer is 34.3% when rounded

percentage of snakes
\frac{# animals}{#snakes} = \frac{100}{?}
**cross multiply 3 and 100 which is 300
<span>**divide 35 into 300</span>
<span>\frac{35}{3}= \frac{100}{?}
</span>the answer is 8.57%
7 0
3 years ago
Find the volume of the largest circular cone that can<br> beinscribed in a shpere of radius 3.
Aleksandr [31]

Answer:

V =\dfrac{32}{3}\pi

Step-by-step explanation:

given,

radius of sphere = 3

volume of cone:

V = \dfrac{1}{3}\pi r^2h

r is the radius of circular base

h is the height of the cone

here r = x and h = 3 + y

now, volume in term of x and y

V = \dfrac{1}{3}\pi x^2(3+y)

Applying Pythagoras theorem

x² + y² = 3²

x = \sqrt{9-y^2}

V = \dfrac{1}{3}\pi ( \sqrt{9-y^2})^2(3+y)

V = \dfrac{1}{3}\pi ( 9-y^2)(3+y)

V = \dfrac{1}{3}\pi (27 + 9 y - 3 y^2-y^3)

differentiating both side

\dfrac{dV}{dy} =\dfrac{1}{3}\pi ( 9-6y- 3y^2)

for maxima  \dfrac{dV}{dy} = 0

\pi ( 3-2 y - y^2)=0

 y² + 2 y - 3 = 0

(y+3)(y-1)=0

 y = 1,-3

y cannot be negative so, volume at y = 1

V = \dfrac{1}{3}\pi (27 + 9 (1)- 3(1)^2-(1)^3)

V =\dfrac{32}{3}\pi

Hence, the largest cone which can be inscribed in the spheres of the radius 3 has volume  V =\dfrac{32}{3}\pi

5 0
3 years ago
If f(x) = 3x - 1 and g(x) = x + 2, find (f - g)(x).
Marina86 [1]

Answer:

\boxed{\sf (f - g)(x) = 2x - 3}

Given:

f(x) = 3x - 1

g(x) = x + 2

To Find:

(f - g)(x)

Step-by-step explanation:

\sf (f -g)(x) = f(x) - g(x) \\ \\   \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \: =(3x - 1) - (x + 2) \\  \\ \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \: =3x - 1 - x - 2 \\  \\ \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \: =3x - x - 1 - 2 \\  \\ \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \: =(3x - x) - (1 + 2) \\  \\ \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \: =2x - (1 + 2) \\  \\ \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \: =2x - 3

6 0
3 years ago
What is a co-prime number? Please answer as soon as possible Thank you
ipn [44]

Answer:

\boxed{\mathrm{view \: explanation}}

Step-by-step explanation:

Two numbers that only have 1 as a common factor are called co-prime numbers.

<u>Example:</u>

Factors of 3 ⇒ 1, 3

Factors of 4 ⇒ 1, 2, 4

These numbers only have 1 as a common factor. So 3 and 4 are co-prime numbers.

6 0
3 years ago
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