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Artemon [7]
3 years ago
9

Scientific Notation is made up of two number parts. The second part is a power of base 10.

Mathematics
1 answer:
Aneli [31]3 years ago
7 0

Answer:

Step-by-step explanation:

True

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Please I need it asap
Sauron [17]

Answer:

5) √(3² + 5²) = √34

6) √(4² - 3²) = √7

7) the visible answers all say exactly the same thing in different ways and are all correct.

Step-by-step explanation:

3 0
3 years ago
Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive
VikaD [51]
Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}

recall that 

\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}

therefore, let's just plug that on the remaining ones,

\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}

now, let's rationalize the denominator on tangent and secant,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
3 0
3 years ago
HELP REALLY NEED HELP PLS :D WINNER GETS 8 POINTS
andrew-mc [135]
The answer to this question is:

280ft^2 for surface area

The volume is 300ft^2


4 0
3 years ago
Compute 7-9-2+10-4?
alexandr1967 [171]
I believe it would be 2. Since 7-9= -2 then subtract another 2 which equals -4 then add 10 which equals 6 then subtract 4. Which equals 2.

2 is the answer, hope this helps

3 0
3 years ago
What is the radius of a circle with the equation x^2+y^2+6x-2y+3?
atroni [7]

Answer:

r = √13

Step-by-step explanation:

Starting with x^2+y^2+6x-2y+3, group like terms, first x terms and then y terms:  x^2 + 6x            + y^2 -2y             = 3.  Please note that there has to be an " = " sign in this equation, and that I have taken the liberty of replacing " +3" with " = 3 ."  

We need to "complete the square" of x^2 + 6x.  I'll just jump in and do it:  Take half of the coefficient of the x term and square it; add, and then subtract, this square from x^2 + 6x:     x^2 + 6x  + 3^2 - 3^2.  Then do the same for y^2 - 2y:  y^2 - 2y + 1^2 - 1^2.

Now re-write the perfect square x^2 + 6x + 9 by (x + 3)^2.  Then we have x^2 + 6x + 9 - 9; also y^2 - 1y + 1 - 1.  Making these replacements:

(x + 3)^2 - 9 + (y - 1)^2 -1 = 3.  Move the constants -9 and -1 to the other side of the equation:  (x + 3)^2 + (y - 1)^2 = 3 + 9 + 1 = 13

Then the original equation now looks like (x + 3)^2 + (y - 1)^2 = 13, and this 13 is the square of the radius, r:  r^2 = 13, so that the radius is r = √13.


8 0
2 years ago
Read 2 more answers
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