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3 years ago

7

One lap around the track is 400 meters.if you run 10 laps around the track,how many miles would you run?Round your answer to the

nearest half-mile
1mile=1609.34meters. (HURRY I NEED THE ANSWER)

1 answer:

photoshop1234 [79]3 years ago

3 0

I got 2.5 miles as my answer do you want to know how I got it...

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Bond [772]

Answer:

$\sin A =\frac{4}{5}$

Step-by-step explanation:

Recall the mnemonics SOH CAH TOA.

SOH means the Sine ratio is Opposite over Hypotenuse.

This implies that;

$\sin A =\frac{Opposite}{Hypotenuse}$

The side opposite to angle A is 4cm and the hypotenuse is 5cm.

$\therefore \sin A =\frac{4}{5}$

6 0

3 years ago

Of the 28 students in class, about 65% are wearing sneakers. How many students are not wearing sneakers? Round to the nearest wh

BigorU [14]

Answer:

10

Step-by-step explanation:

Based on the given conditions, formulate: \(28 \times ( 1 - 65\% )\)

Calculate the sum or difference:\(28 \times 0.35\)

Calculate the product or quotient:\(9.8\)

Find the closest integer: 10

Answer: 10

7 0

2 years ago

1. Mrs. Verner's class has

Gemiola [76]

Answer:

46.7%

Step-by-step explanation:

Given:

Total number of students in Mrs. Verner's class = 15

Number of girls = 8

To find: percentage are boys

Solution:

Percentage of boys = ( Number of boys / Total number of students ) × 100

Number of boys = Total number of students - Number of girls = 15 - 8 = 7

So,

Percentage of boys = × 100 = 46.7%

4 0

2 years ago

Please help me up there

Anettt [7]

The answer is 38,400

The scale is 1:40 so the side that is 4 in becomes 160 and the side that is 6 in becomes 240
Multiply 160 and 240 and that is the area of the room

Hope this helps

6 0

3 years ago

Given that f(x) = x + 3<br> a) Find f(2)

Nataly_w [17]

${ \qquad\qquad\huge\underline{{\sf Answer}}}$

Here we go ~

In the above question, it is given that :

$\qquad \sf \boxed{ \sf f(x) = \frac{x + 3}{2} }$

A.) Find f(2) :

$\qquad \sf \dashrightarrow \: f(2) = \dfrac{2 + 3}{2}$

$\qquad \sf \dashrightarrow \: f(2) = \dfrac{5}{2}$

or

$\qquad \sf \dashrightarrow \: f(2) = 0.5$

B.) Find ${ \sf {f}^{-1}(x) }$ :

$\qquad \sf \dashrightarrow \: let \: y = f (x)$

so, we can write it as :

$\qquad \sf \dashrightarrow \: y = \dfrac{x + 3}{2}$

$\qquad \sf \dashrightarrow \: 2y = x + 3$

$\qquad \sf \dashrightarrow \: x = 2y - 3$

Now, put x = ${ \sf {f}^{-1}(x) }$ , and y = x and we will get our required inverse function ~

$\qquad \sf \dashrightarrow \: f {}^{ - 1}(x) = 2x- 3$

C.) Find ${ \sf {f}^{-1}(12) }$ :

$\qquad \sf \dashrightarrow \: f {}^{ - 1}(12) = 2(12)- 3$

$\qquad \sf \dashrightarrow \: f {}^{ - 1}(12) = 36- 3$

$\qquad \sf \dashrightarrow \: f {}^{ - 1}(12) = 33$

7 0

1 year ago