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d1i1m1o1n [39]
3 years ago
13

On a separate piece of paper, draw the different ways you can divide the given shape into simpler shapes. Describe each result.

Mathematics
1 answer:
marishachu [46]3 years ago
8 0

Answer:

there will be 4 cornered shapes 1 rectangle that is almost a square and the shape I Forgot TvT

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What is the mean of these numbers?4,5,5,2,3,3,2,8
VladimirAG [237]

Answer:

4

Step-by-step explanation:

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3 years ago
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There's a one time fee of $27 to join the gym the monthly cost of using the gym is $18 write an equation for the relationship th
Sonbull [250]

y = 18x + 27
with y equaling the total and x equaling the number if months
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In the diagram below, LATE is an isosceles trapezoid with LE ≅ AT , LA = 24, ET = 40, and AT = 10. Altitudes LF and AG are drawn
weqwewe [10]

Answer:

6

Step-by-step explanation:

From the Trapezoid attached :

EF = GT

FG = LA

LE = AT = 10

LA = 24 ; FG = 24

FG + EF + GT = 40

Let : EF and GT = x

FG + 2x = 40

24 + 2x = 40

2x = 40 - 24

2x = 16

x = 16 ÷ 2 = 8

Hence, EF = GT = 8

Using Pythagoras :

Opposite² = hypotenus² - Adjacent²

LF² = LE² - FE²

LF² = 10² - 8²

LF² = 100 - 64

LF² = 36

LF = √36

LF = 6

4 0
3 years ago
Question 101 points)
maw [93]

Answer:

\large\boxed{y=-\dfrac{2}{3}x+\dfrac{13}{3}}

Step-by-step explanation:

\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\\\m-slope\\b-y-intercept\\\\\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\===============================

\text{We have the point:}\\\\(5,\ 1)\ \text{and}\ (-4,\ 7).\ \text{Substitute:}\\\\m=\dfrac{7-1}{-4-5}=\dfrac{6}{-9}=-\dfrac{6:3}{9:3}=-\dfrac{2}{3}\\\\\text{We have the equation in form:}\\\\y=-\dfrac{2}{3}x+b\\\\\text{Put the coordinates of the point (5, 1) to the equation:}\\\\1=-\dfrac{2}{3}(5)+b\\\\1=-\dfrac{10}{3}+b\qquad\text{add}\ \dfrac{10}{3}\ \text{to the both sides}\\\\\dfrac{3}{3}+\dfrac{10}{3}=b\to b=\dfrac{13}{3}\\\\\text{Finally:}\\\\y=-\dfrac{2}{3}x+\dfrac{13}{3}

4 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
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