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sweet-ann [11.9K]
3 years ago
11

Y=2x 3 help answer plz summer packet

Mathematics
1 answer:
MrMuchimi3 years ago
7 0
The answer what I think is 6x
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(9-3)+18÷8 <br>show your work ​
egoroff_w [7]

Answer:

(9-3)+18÷8

=6+ 9/4

Step-by-step explanation:

.............

3 0
2 years ago
Read 2 more answers
Please I need hep, get free brainly :D
padilas [110]

Step-by-step explanation:

<u>Formula:</u>

\boxed{a = 3.14(r)^{2} }

r represents the radius.

The radius is 1/2 of the diameter.

Since 1/2 of 16 is 8, 8 is our radius.

<u>Substitute:</u>

a = 3.14(8)^{2}

<u>Solve:</u>

a = 3.14(64)

a = 200.96in^{2}

6 0
2 years ago
Read 2 more answers
Heelppppppppp!!!!! Please?!
Gre4nikov [31]

To calculate h evaluate inside the square root before taking the square root

h = √(7² - 3.5²) = √(49 - 12.25) = √36.75 = 6.06 ( to 2 dec. places)


3 0
3 years ago
Read 2 more answers
Video Example EXAMPLE 1 Find the linearization of the function f(x) = x + 1 at a = 3 and use it to approximate the numbers 3.98
Blizzard [7]

Answer:

  the linearization is y = 1/4x +5/4

  the linearization will produce <em>overestimates</em>

  the values computed from this linearization are ...

     f(3.98) ≈ 2.245

     f(4.05) ≈ 2.2625

Step-by-step explanation:

Apparently, you have ...

  f(x)=\sqrt{x+1}

from which you have correctly determined that ...

  f'(x)=\dfrac{1}{2\sqrt{x+1}}

so that f(3) = 2 and f'(3) = 1/4. Putting these values into the point-slope form of the equation of a line, we get the linearization ...

  g(x) = (1/4)(x -3) +2

  g(x) = (1/4)x +5/4

__

The values from this linearization will be overestimates, as the curve f(x) is concave downward everywhere. The tangent (linearization) is necessarily above the curve everywhere.

__

At the given values, we find ...

  g(3.98) = 2.245

  g(4.05) = 2.2625

4 0
3 years ago
Which angle has a sine of -1/2 and a cosine of -√3/2
Sergeeva-Olga [200]
\sin x=-\dfrac{1}{2} < 0\\\\\cos x=-\dfrac{\sqrt3}{2} < 0\\\\therefore\ x\in(180^o;\ 270^o)

\text{We know:}\ \tan x=\dfrac{\sin x}{\cos x}\\\\\text{therefore:}\ \tan x=\dfrac{-\frac{1}{2}}{-\frac{\sqrt3}{2}}=\dfrac{1}{2}\cdot\dfrac{2}{\sqrt3}=\dfrac{1}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=\dfrac{\sqrt3}{3}

\tan x=\dfrac{\sqrt3}{3}\Rightarrow x=30^o+180^o\cdot k;\ k\in\mathbb{Z}

x\in(180^o;\ 270^o)\ therefore\ \boxed{x=30^o+180^o=210^o}
8 0
3 years ago
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