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melomori [17]
3 years ago
9

when tom simplified the expression -2.6 + (-5.4), he got 2.8 what mistake did tom likely make? pls help me

Mathematics
1 answer:
olga_2 [115]3 years ago
5 0

Answer:

2.6 and 5.4 are both the same sign, he should add them and then take that negative sign from (-5.4), so the answer would be -8.0.

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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
3c+8=14 linear equation solve for variable​
slava [35]

Answer:

c=2

Step-by-step explanation:

3c+8=14

subtract 8 from both sides

3c=6

divide both sides by 3

c=2

3 0
3 years ago
What is the area of this rectangle in square inches??
rodikova [14]
The area is A. 84 square inches
7 0
3 years ago
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Combine the radicals. 3√5-8√5+2√5
weeeeeb [17]

Answer:  -3\sqrt{5}

-3 times the square root of 5

=============================================

Explanation:

Let x = \sqrt{5}

Replace all the root 5 terms with x and we go from

3\sqrt{5}-8\sqrt{5}+2\sqrt{5}

to

3x-8x+2x

From here, combine like terms to get

3x-8x+2x = -5x+2x = -3x

and the last thing to do is replace the x with sqrt(5)

-3x = -3\sqrt{5}

Meaning that,

3x-8x+2x = -3x

3\sqrt{5}-8\sqrt{5}+2\sqrt{5} = -3\sqrt{5}

7 0
3 years ago
Five times the difference of a number and twelve is greater than forty.
Naya [18.7K]

Answer:

This is the value in numbers.

5(n-12)>40

Step-by-step explanation:

Let's solve your inequality step-by-step.

5(n−12)>40

Step 1: Simplify both sides of the inequality.

5n−60>40

Step 2: Add 60 to both sides.

5n−60+60>40+60

5n>100

Step 3: Divide both sides by 5.

5n

5

>

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n>20

Answer:

n>20

6 0
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