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bekas [8.4K]
3 years ago
5

Wich number does not represent a whole number

Mathematics
1 answer:
erica [24]3 years ago
7 0
C. 20/4 because a fraction isn’t a whole number.
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I don't know how to do this. I forget plz help
Leno4ka [110]
Ok so 
0.5%=0.005
0.005x=8
x=8/0.005
x=1600
8 is 0.5% of 1600
Hope this helps
If u still don't understand plz message me
If u understand plz brainlest me
If you continue with the same formula with a different problem then you will get it correct 
Thank you!
3 0
3 years ago
Read 2 more answers
What is the constant variation in the equation 3y=6x? Explain. PLEASE HELP IM DESPERATE
lbvjy [14]
Hi Desperate!

anyways the equation for constant variation is y=kx
3y=6x is it, but we don't want the 3.
so we divide on both sides to cancel out the 3.
y=2x
4 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
(7^2)^4= n^8 solve for n
S_A_V [24]

Answer:

n=7

Step-by-step explanation:

(7^2)^4= n^8

We know that a^b^c = a^(b*c)

7^(2*4) = n^8

7^8 = n^8

Since the exponents are the same, the bases must be the same

n=7

4 0
3 years ago
Solve the equation for a K = 4a + 9ab
Nuetrik [128]

Answer:

K/(4+9b) = a

Step-by-step explanation:

K = 4a + 9ab

Factor out a on the right hand side

K = a(4 + 9b)

Divide each side by (4+9b)

K/(4+9b) = a(4 + 9b)/(4+9b)

K/(4+9b) = a

3 0
3 years ago
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