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kirill [66]
3 years ago
15

F(x) is a function. 3 5 2 10 15 0 f(x) O A. True O B. False

Mathematics
2 answers:
elixir [45]3 years ago
3 0

Answer:

False

Step-by-step explanation:

X = 5 goes to two different values of y so it is not a one to one correspondence.  This is a relation not a function.

Dmitry_Shevchenko [17]3 years ago
3 0
The answer is false, 5 goes to 2 different numbers
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Sonja [21]

I "estimate" 600....?


7 0
3 years ago
Read 2 more answers
Help 15 points!!!!!!!!!!!!!!!!!!!!!
Svetllana [295]

Answer:

see explanation

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan A = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{7}{8} , then

∠ A = tan^{-1} (\frac{7}{8} ) ≈ 41° ( to the nearest degree )

The sum of the angles in the triangle = 180° , then

∠ B + 41° + 90° = 180°

∠ B + 131° = 180° ( subtract 131° from both sides )

∠ B = 49°

Using Pythagoras' identity in the right triangle

AB² = BC² + AC² = 7² + 8² = 49 + 64 = 113 ( take square root of both sides )

AB = \sqrt{113} ≈ 10.6 ( to the nearest tenth )

3 0
2 years ago
Differential cos^2x dy/dx =e^y-tanx​
gulaghasi [49]

Answer:

y=t−1+ce

−t

where t=tanx.

Given, cos

2

x

dx

dy

+y=tanx

⇒

dx

dy

+ysec

2

x=tanxsec

2

x ....(1)

Here P=sec

2

x⇒∫PdP=∫sec

2

xdx=tanx

∴I.F.=e

tanx

Multiplying (1) by I.F. we get

e

tanx

dx

dy

+e

tanx

ysec

2

x=e

tanx

tanxsec

2

x

Integrating both sides, we get

ye

tanx

=∫e

tanx

.tanxsec

2

xdx

Put tanx=t⇒sec

2

xdx=dt

∴ye

t

=∫te

t

dt=e

t

(t−1)+c

⇒y=t−1+ce

−t

where t=tanx

8 0
2 years ago
What is the square root of 9/16, 9 to the power of two
lbvjy [14]

Answer:

\frac{3}{4} , 9

Step-by-step explanation:

\sqrt{9/16} = \sqrt{9} ÷ \sqrt{16} = \frac{3}{4}

\sqrt{9^2} = \sqrt{81} = 9


3 0
3 years ago
Given f(x) = 2x + 3 and g(x) = x - 9, what is f(x) + g(x)?
Orlov [11]
Solution:

f(x) = 2x + 3
g(x) = x - 9

f(x) + g(x)
= (2x + 3) + (x - 9)
= 2x + x + 3 - 9
= 3x - 6
6 0
3 years ago
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