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2 years ago

Pleaseeeee help!!!!!!!!

Mathematics

2 answers:

ELEN [110]2 years ago

3 0

Answer:

Step-by-step explanation:

The total number of people is

5+6+4+7+5+3 = 30 people

devlian [24]2 years ago

3 0

Answer:

Step-by-step explanation:

pizza:5+7=12

Burger:6+5=11

Other:4+3=7

Total:12+11+7=30

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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,

$\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}$

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2 years ago

Janet has 17 quarters and $13 in bills. How much total money does she have?

galben [10]

The answer is $17.25 because you do 17 divided by 4 since 4 quarters goes into a dollAR

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2 years ago

The vertices of figure BCDE have coordinates B(4, 4), C(−4, 4), D(−4, −4), and E(4, −4). The vertices of figure BꞌCꞌDꞌEꞌ have co

nikdorinn [45]

C.) A <span>dilation with a scale factor of 4, with center of dilation at the origin

Here, You can either draw the figure or compare the coordinates,
B = (4, 4) then B' = (1, 1)
C = (-4, 4) then C' = (-1, 1)
D = (-4, -4) then D' = (-1, -1)
E = (4, -4) then E' = (1, -1)

So, Every coordinate is reducing by 4th factor... Hence, dilation factor is 4 about the origin

Hope this helps!</span>

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2 years ago

Dimitri drew a right angle. Then he drew an angle that was 20° larger. What is the measure of the second angle Dimitri drew?

marysya [2.9K]

Right angle is 90 degrees. 90 + 20 = 110 degrees

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2 years ago

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Write the common rule to describe the translation using coordinate notation.

UNO [17]

The common rule is (x - 1, y + 3) can be used to describe the translation.

Step-by-step explanation:

Step 1:

The point J (-3, -4) becomes $J^{1}$ (-4, -1).

In order to write the rule for translation from J to $J^{1}$ , we subtract the x coordinate of J from $J^{1}$ and subtract the y coordinate of J from $J^{1}$ .

The x coordinate $= -4-(-3) = -4+3 = -1.$

The y coordinate $= -1-(-4) = -1+4 = 3.$

Step 2:

So from the calculations, we get that the x coordinate is subtracted by 1 i.e. $x-1$ and the y coordinate is increased by 3 i.e. $y+3$ .

So the common rule is (x - 1, y + 3).

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2 years ago