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Misha Larkins [42]
3 years ago
14

What’s the length of BD ?

Mathematics
2 answers:
Sergeeva-Olga [200]3 years ago
5 0
The answer is 4.99.

Since tan = oposite/adjacent we can plug in the values we know.

Rina8888 [55]3 years ago
3 0

Answer:

BD = 4.99

Step-by-step explanation:

You can simply use the trigonometric identity tangent to solve for length BD.

Tan = opposite/adjacent

In this case we have,

Tan 31 = 3/BD

BD = 3/Tan 31

BD = 4.99

You might be interested in
What is<br> 5 1/6 - 5 1/2
Ymorist [56]

Answer:

1/3

Step-by-step explanation:

51/6 = 51/6

51/2= 5 3/6

= 2/6

6 0
3 years ago
Read 2 more answers
THIS IS A MULTI CHOICE ANSWER
Paha777 [63]
<span>C. 
-2/3

</span><span>A. 
-1/2

derp derp</span>
4 0
3 years ago
Plz help!
beks73 [17]

First, you need to find the derivative of this function.  This is done by multiplying the exponent of the variable by the coefficient, and then reducing the exponent by 1.  

f'(x)=3x^2-3

Now, set this function equal to 0 to find x-values of the relative max and min.

0=3x^2-3

0=3(x^2-1)

0=3(x+1)(x-1)

x=-1, 1

To determine which is the max and which is the min, plug in values to f'(x) that are greater than and less than each.  We will use -2, 0, 2.

f'(-2)=3(-2)^2-3=3(4)-3=12-3=9

f'(0)=3(0)^2-3=3(0)-3=0-3=-3

f'(2)=3(2)^2=3(4)-3=12-3=9

We examine the sign changes to determine whether it is a max or a min.  If the sign goes from + to -, then it is a maximum.  If it goes from - to +, it is a minimum.  Therefore, x=-1 is a relative maximum and x=1 is a relative miminum.

To determine the values of the relative max and min, plug in the x-values to f(x).

f(-1)=(-1)^3-3(-1)+1=-1+3+1=3

f(1)=(1)^3-3(1)+1=1-3+1=-1

Hope this helps!!

7 0
3 years ago
.
bekas [8.4K]

Answer:

It's a line

Step-by-step explanation:

A line is not a three dimensional shape. I hope this helps :) also let me know if you got it right!

5 0
3 years ago
The distance d of a particle moving in a straight line is given by d(t) = 2t3 + 5t – 2, where t is given in seconds and d is mea
Sergeeva-Olga [200]

Answer:

(C)6t^2+5

Step-by-step explanation:

Given the distance, d(t) of a particle moving in a straight line at any time t is:

d(t) = 2t^3 + 5t - 2, $ where t is given in seconds and d is measured in meters.

To find an expression for the instantaneous velocity v(t) of the particle at any given point in time, we take the derivative of d(t).

v(t)=\dfrac{d}{dt}\\\\v(t) =\dfrac{d}{dt}(2t^3 + 5t - 2) =3(2)t^{3-1}+5t^{1-1}\\\\v(t)=6t^2+5

The correct option is C.

6 0
3 years ago
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