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nirvana33 [79]
3 years ago
10

Angel and Jayden were at track practice. The track is kilometers around.

Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
6 0

Answer:

Jayden is running faster

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815 dived by 15 in quotiens
Dmitriy789 [7]
Your answer will be a repeating decimal or fraction

54.333

54 1/3
4 0
3 years ago
Plzzzz help me to find the answer
Natalka [10]

Answer:

im pretty sure its 67, sorry if I'm wrong

7 0
3 years ago
(10.04 MC)
Luba_88 [7]

Answer:

f

(

x

)

=

3

cos

(

2

x

)

+

2

EXPLANATION:

We are able to use the transformation formula

f

(

x

)

=

a

⋅

cos

(

x

−

h

b

)

+

k

. You start with

f

(

x

)

=

cos

(

x

)

and replace

a

with the desired amplitude,

h

with the desired horizontal shift, and

k

with the desired vertical shift. This leaves out the

b

-value. A regular cosine function has a period of

2

π

. If you want a period of

π

, since that is one half of the original period, you need to replace your

b

with a

1

2

.

This is about how it would work out.

f

(

x

)

=

3

⋅

cos

(

x

−

0

1

2

)

+

2

From there you simplify your equation giving you

f

(

x

)

=

3

cos

(

2

x

)

+

2

4 0
3 years ago
Read 2 more answers
For questions 8 – 14, use the following functions:
pogonyaev

we are given

f(x)=5

g(x)=\sqrt{x+2}

(8)

(f+g)(x)=f(x)+g(x)

we can plug it

(f+g)(x)=5+\sqrt{x+2}

(9)

(f-g)(x)=f(x)-g(x)

we can plug it

(f-g)(x)=5-\sqrt{x+2}

(10)

(f*g)(x)=f(x)*g(x)

we can plug it

(f*g)(x)=5\sqrt{x+2}

(11)

(\frac{f}{g} )(x)=\frac{f(x)}{g(x)}

we can plug it

(\frac{f}{g} )(x)=\frac{5}{\sqrt{x+2}}

(12)

(\frac{g}{f} )(x)=\frac{g(x)}{f(x)}

we can plug it

(\frac{g}{f} )(x)=\frac{\sqrt{x+2}}{5}

(13)

(fog)(x)=f(g(x))

f(x)=5

we can replace g(x)

we get

(fog)(x)=5

(14)

(gof)(x)=g(f(x))

f(x)=5

we can replace f(x)

(gof)(x)=\sqrt{f(x)+2}

we get

(gof)(x)=\sqrt{5+2}

(gof)(x)=\sqrt{7}


8 0
3 years ago
Solve equation. 2x+5=5+2x
ivann1987 [24]

2x + 5 = 5 + 2x

2x - 2x = 5 - 5

0 = 0

true

--------------------

5 0
3 years ago
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