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Dominik [7]
3 years ago
8

Evaluate the expression 2m+2n if m=9 and n=3

Mathematics
2 answers:
tatuchka [14]3 years ago
4 0
Just substitute each number in for the variables: 9 for m and 3 for n.

2m+2n
= 2(9)+2(3)
= 18+6
= 24
VashaNatasha [74]3 years ago
4 0
Substitute m and n values into the expression.
2(9)+2(3)
18+6= 24
24 is your answer.
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Simplify the ratio 19:114
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\frac{19}{114}=\frac{19}{19 \cdot 6}=\frac{1}{6}
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The ratio of green houses to white houses in Queens is 5 to 7. There are 63 white houses, How many green houses are there?
Ludmilka [50]
45 to 63

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3 0
3 years ago
Divide € 165 into the ratio 1:2:12
fenix001 [56]

Answer:

11 : 22 : 132

Step-by-step explanation:

1 + 2 + 12 = 15

165 ÷ 15 = 11

1 × 11 = 11

2 × 11 = 22

12 × 11 = 132

11 : 22 : 132

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
Whats the equation of the line that passes threw -4,4 and is parrell to the line y=1/2-4
vovangra [49]

Answer:

y=\frac{1}{2}x+6

Step-by-step explanation:

Given:

The equation of the known line is:

y=\frac{1}{2}x-4

A point on the unknown line is (-4, 4)

Now, since the two lines are parallel, their slopes must be equal.

Now, slope of the known line is the coefficient of 'x' which is \frac{1}{2}.

Therefore, the slope of the unknown line is also m=\frac{1}{2}

Now, for a line with slope 'm' and a point on it (x_1,y_1) is given as:

y-y_1=m(x-x_1)

Here, m=\frac{1}{2}, x_1=-4,y_1=4. Therefore,

y-4=\frac{1}{2}(x-(-4))\\\\y-4=\frac{1}{2}(x+4)\\\\y-4=\frac{1}{2}x+2\\\\y=\frac{1}{2}x+2+4\\\\y=\frac{1}{2}x+6

Hence, the equation of the unknown line is y=\frac{1}{2}x+6.

5 0
4 years ago
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