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vampirchik [111]
3 years ago
5

Ida paid 75¢ for two iteams and got 19¢ in change. What iteams did she buy

Mathematics
2 answers:
qaws [65]3 years ago
5 0
You could buy the Magnet and the Postcard and get 19 cents!
Ronch [10]3 years ago
3 0
She brought the magnets i think
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How many solutions does an equation of the form x^2- k^2= 0 have?
N76 [4]
1. Use Difference of squares: a^2-b^2=(a+b)(a-b)
(x+k)(x-k)=0

2. Solve for x and k
x=+-k
k=+-x
8 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
Plx help agajb lsksosjb​
SashulF [63]

Answer:

the third one

Step-by-step explanation:

4 0
3 years ago
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The ratio of boys and teacher in a school is 2:3 the ratio of teachers to the girls is 4:5.Find the ratio of all the three in a
Scorpion4ik [409]

Answer:

Step-by-step explanation:

2:3 boy:teacher 8:12

4:5 teacher:girl 12:15

8:12:15 boy:teacher:girl

7 0
2 years ago
461 divided by 3 place the digit
Gekata [30.6K]
I believe it's 153.. 
3 0
3 years ago
Read 2 more answers
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