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natulia [17]
3 years ago
12

6 + x when x equals 3

Mathematics
2 answers:
Drupady [299]3 years ago
8 0

Answer:

6 + x= when x equals 3. If x equals 3 then the equations is written as 6+3 which 9.

The answer is 9.

Step-by-step explanation:

MissTica3 years ago
4 0

Answer: the answer is 9 duhh

Step-by-step explanation:

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K=mv^2/2<br> Find derivative of k
jonny [76]

Answer: I think. it's 0

Step-by-step explanation:

5 0
3 years ago
Identify the vertex of g(x) = (x+10)^2+2
RoseWind [281]

Answer:

(-10, 2)

Step-by-step explanation:

Compare:

g(x) = (x+10)^2+2

f(x)  = (x - h)^2 + k

Matching these up, term by term, we see that h = -10 and k = 2.  (h, k) represents the vertex of the parabola f(x)  = (x - h)^2 + k.

We conclude that the vertex of this parabola is (-10, 2).

7 0
3 years ago
One angle of a triangle measures 95°. The other two angles are in a ratio of 8:9. What are the measures of those two angles?
Luba_88 [7]

Answer:

40 degrees and 45 degrees

Step-by-step explanation:

If you subtract 95 degrees from a 180-degree triangle, you are left with 85 degrees. To proportionally split them, find two values that add to the same value. Since 40 and 45 make up 80 and they multiply by two to make 80:90, these are the two measures.

8 0
3 years ago
Find the area of a regular octagon with an apothem of 8.45 cm and a side length of 7 cm. (round to the nearest whole number) A)
Alexeev081 [22]

Answer:

237

Step-by-step explanation:

5 0
3 years ago
An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum valu
exis [7]
Take the derivative with respect to t
- w \sin(wt) + \sqrt{8} w cos(wt)
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
0 = -w \sin(wt) + \sqrt{8} w cos(wt)
divide by w
0 =- \sin(wt) + \sqrt{8} cos(wt)
we add sin(wt) to both sides

\sin(wt)= \sqrt{8} cos(wt)
divide both sides by cos(wt)
\frac{sin(wt)}{cos(wt)}= \sqrt{8}   \\  \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\  \\ wt=arctan(2 \sqrt{2)} OR\\ wt=arctan( { \frac{1}{ \sqrt{2} } )
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

I=cos(2n \pi -2arctan( \sqrt{2} ))
since 2npi is just the period of cos
cos(2arctan( \sqrt{2} ))= \frac{-1}{3} &#10;
substituting our second soultion we get
I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))
since 2npi is the period
I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}
so the maximum value =\frac{1}{3}
minimum value =- \frac{1}{3}


4 0
3 years ago
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