Help me please with this

Mathematics

1 answer:

prisoha [69]3 years ago

6 0

To factor this collect the like terms..2y would go with -7y to make -5y etc. Then factor the expression by factoring out 9 Your final answer should be 9(1-y+x^2)

If you need to simplify this calculate the sum or difference between numbers (3-1+7=9)
Your final answer for this should be 9-9y+9x^2

Hope this helps!

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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

Please help!! How do you do number 1?

alisha [4.7K]

Do a ratio tall over shadow. How tall is tree / shadow. How tall is person / shadow then cross multiply. X/7.5 =5/3. 37.5=3x. Divide by 3 answer 12.5

6 0

3 years ago

-6f+13=2f-11

Paha777 [63]

The answer is f=3. You get this by combining like terms.

7 0

3 years ago

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A student took a total of 4 tests over the course of 8 weeks. How many weeks of school will the student attend to take a total o

Julli [10]

Answer:

4/8 = 20/x

20 x 8 = 160

160 ÷ 4 = 40

the student will attend 40 weeks of school

3 0

3 years ago

How many 6-digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, if repetitions of digits are allowed?

sveta [45]

There are 6 digits. Each digit can take ten different numbers except for the first digit since it cannot be zero.

So:

9 x 10 x 10 x 10 x 10 x 10

900000 numbers.

Another way of thinking about this is to just count up to 999,999. Obviously there are 999,999 different numbers here. But since our number has to have 6 digits in them, we have to delete 99,999 numbers. Thus there are 900,000 different numbers.

6 0

3 years ago