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xz_007 [3.2K]
3 years ago
11

Can you please help me with the questions I will give you brainliest.

Mathematics
1 answer:
KiRa [710]3 years ago
8 0

Answer:

C the First page

Step-by-step explanation:

1) you input the 2 into the T So the equation would look like this

15+2.5*(times) 2

2) I'm not sure for the second

3) A because you are using the number given then multiplying.

1.58* times 5 = 7.5

But the starting number would be 1.5 sense that's  the first number given

No one report me please.

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Is -2.7 an irrational number
ololo11 [35]

Answer:

No

Step-by-step explanation:

An irrational number is a number that never ends or repeats. This decimal ends and it doesn't repeat.

6 0
3 years ago
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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
A van can travel 18 miles on each gallon of gasoline. at that rate how many miles can the van travel on 15 gallons of gasoline
tensa zangetsu [6.8K]
18 / 1 = x / 15......18 miles to 1 gallon = x miles to 15 gallon
cross multiply
(1)(x) = (18)(15)
x = 270 miles
7 0
3 years ago
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Please help me solve this angle problem. It is due today!
Paul [167]
Answer: B
No explanation
7 0
3 years ago
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Dimas [21]

Answer: Jenny

Step-by-step explanation:

By subtitue Y = 0

0=(x-3)^2

And you get 3

3 0
3 years ago
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