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3 years ago

Dana bikes 10 mi in 50 min. At this rate, how far can she bike in 90 min?

Mathematics

2 answers:

lara31 [8.8K]3 years ago

8 0

The answer will be 18 miles

scoray [572]3 years ago

7 0

18 miles in 90 minutes. If you think about it she's going 1 mile every 5 minutes. So if she biked 10 miles in 50 she would have hone 8 miles in 40 minutes. You now simply add 10 miles + 8 miles and you end up with 18 miles.

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If AC = 40, find the length of JK.

lesantik [10]

Answer:

JK = 24

Step-by-step explanation:

Δ BKJ and Δ BCA are similar triangles and ratios of corresponding sides are equal, that is

$\frac{JK}{AC}$ = $\frac{BK}{BC}$ , substitute values

$\frac{JK}{40}$ = $\frac{3}{5}$ ( cross- multiply )

5JK = 120 ( divide both sides by 5 )

JK = 24

8 0

3 years ago

Read 2 more answers

- If f ( x ) = x² + 2x - 1 and g ( x ) = 3x - 2 , then verify the following :

RideAnS [48]

Step-by-step explanation:

Hey there!

Given;

f ( x ) = x² + 2x - 1

g ( x ) = 3x - 2

To verify:

$( \frac{f}{g} )(2) = \frac{f(2)}{g(2)}$

LHS:

$(\frac{f}{g} )(x) = \frac{ {x}^{2} + 2x - 1}{3x - 2}$

~ Insert "2" instead of "x".

$(\frac{f}{g} )(2) = \frac{ {2}^{2} + 2 \times 2 - 1 }{3 \times 2 - 2}$

Simplify it;

$\frac{f}{g} (2) = \frac{4 + 4 - 1}{6 - 2}$

Therefore; (f/g)(2) = 7/4.

RHS:

$\frac{f(x)}{g(x)} = \frac{ {x}^{2} + 2x - 1}{3x - 2}$

~Insert "2" instead of"x".

$\frac{f(2)}{g(2)} = \frac{ {2}^{2} + 2 \times 2 - 1 }{3 \times 2 - 2}$

Simplify it.

$\frac{f(2)}{g(2)} = \frac{4 + 4 - 1}{6 - 2}$

Therefore, f(2)/g(2) = 7/4.

Since (f/g)(2) = f(2)/g(2) = 7/4.

Proved!

Hope it helps!

3 0

3 years ago

Sue Anne owns a medium-sized business. The probability model below describes the number of employees that may call in sick on an

Masja [62]

Answer:

The mean is $\mu = 2.1$

Step-by-step explanation:

The mean number of employees calling sick per day can be mathematically represented as

$\mu = 0* 0.05 + 1 * 0.40 + 2 * 0.3 + 3 * 0.3 + 4*0.05$

$\mu = 2.1$

4 0

3 years ago

A pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm, and 96 cm but does not resonate at any wave

erma4kov [3.2K]

Answer:

A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:

A. closed at both ends

B. open at one end and closed at one end

C. open at both ends.

D. we cannot tell because we do not know the frequency of the sound.

The right choice is:

B. open at one end and closed at one end .

Step-by-step explanation:

Given:

Length of the pipe, $L$ = 120 cm

Its wavelength $\lambda_1$ = 480 cm

$\lambda_2$ = 160 cm and $\lambda_3$ = 96 cm

We have to find whether the pipe is open,closed or open-closed or none.

Note:

The fundamental wavelength of a pipe which is open at both ends is 2L.
The fundamental wavelength of a pipe which is closed at one end and open at another end is 4L.

So,

The fundamental wavelength:

⇒ $4L=4(120)=480\ cm$