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2 years ago

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In her backyard, Mrs. Wilson set up an area for the kids to run races. The race area was 15 yards across. Mrs. Wilson put cones

every ½ yard to mark the location. How many cones did she place in her backyard?

2 answers:

Nostrana [21]2 years ago

7 0

The answer would be 30 cones. How you do it is just divide 15 by 0.5. Then you’ll get 30.

Karo-lina-s [1.5K]2 years ago

5 0

Answer:

get rick rolled

Step-by-step explanation:

You might be interested in

Miya is picking up two friends to go the beach. She drives from her house to

Serga [27]

Answer:

Option (C)

Step-by-step explanation:

Coordinates of the point representing the location of Miya → (1, 10)

Coordinates of the point representing the location of Drea → (13, 1)

Coordinates of the point representing the location of Francine → (16, 1)

Coordinates of the point representing the location of Beach → (19, 4)

Distance between Miya and Drea = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

= $\sqrt{(13-1)^2+(1-10)^2}$

= $\sqrt{144+81}$

= 15 miles

Distance between Drea and Francine = $\sqrt{(16-13)^2+(1-1)^2}$

= 3 miles

Distance between Francine and Beach = $\sqrt{(19-16)^2+(4-1)^2}$

= $\sqrt{9+9}$

≈ 4.2 miles

Total distance between Miya and the beach = 15 + 3 + 4.2

= 22.2 miles

Option (C) is the answer.

5 0

2 years ago

Pls helpPlsplsplsplspls

Vedmedyk [2.9K]

Answer:

it's applied to an force reallllll

4 0

2 years ago

steve has 75.50$ to spend at the movies. His ticket is 45.50? and he gets a large popcorn and soda. If the popcorn and soda each

Maru [420]

$15. 75.5 - 45.5 = 30.

30/2 = 15

5 0

2 years ago

Twenty planes are at an airport. Eight planes take off, then five planes land, then 7

steposvetlana [31]

Answer:

19

Step-by-step explanation:

20 - 8 = 12

12 + 5 = 17

17 - 7 = 10

10 + 9 = 19

4 0

3 years ago

Which pair of funtions is not a pair of inverse functions? please help!!

antiseptic1488 [7]

Answer:

$f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Step-by-step explanation:

we know that

To find the inverse of a function, exchange variables x for y and y for x. Then clear the y-variable to get the inverse function.

we will proceed to verify each case to determine the solution of the problem

<u>case A)</u> $f(x)=\frac{x+1}{6} , g(x)=6x-1$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y+1}{6}$

Isolate the variable y

$6x=y+1$

$y=6x-1$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=6x-1$

therefore

f(x) and g(x) are inverse functions

<u>case B)</u> $f(x)=\frac{x-4}{19} , g(x)=19x+4$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y-4}{19}$

Isolate the variable y

$19x=y-4$

$y=19x+4$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=19x+4$

therefore

f(x) and g(x) are inverse functions

<u>case C)</u> $f(x)=x^{5}, g(x)=\sqrt[5]{x}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=y^{5}$

Isolate the variable y

fifth root both members

$y=\sqrt[5]{x}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\sqrt[5]{x}$

therefore

f(x) and g(x) are inverse functions

<u>case D)</u> $f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y}{y+20}$

Isolate the variable y

$x(y+20)=y$

$xy+20x=y$

$y-xy=20x$

$y(1-x)=20x$

$y=20x/(1-x)$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=20x/(1-x)$

$\frac{20x}{1-x}\neq \frac{20x}{x-1}$

therefore

f(x) and g(x) is not a pair of inverse functions

7 0

2 years ago

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