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earnstyle [38]
3 years ago
12

Chang knows one side of a triangle is 13 cm. Which set of two sides is possible for the lengths of the other two sides

Mathematics
2 answers:
ladessa [460]3 years ago
7 0
8 cm & 9 cm .. hope that helps u out :)
cupoosta [38]3 years ago
3 0

Answer:

8 cm and 9 cm

Step-by-step explanation:

In order to find out, you add both 8 and 9 together. If the sum is above/higher than the number 13, they can all be made into a triangle. 8+9 is 17 and 17 is higher than 13.

hope this helps!

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vesna_86 [32]

Answer:

2/5 (x-1)=15 (x=7)

7 (1+3m)=49 (m=2)

6 0
3 years ago
Read 2 more answers
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
Will give brainliest lots of points A is the midpoint of LB,Lb =6x-17 and AB=2x+3
NARA [144]
A is the mid point ==> AL = AB = 2AB
6x-17=2(2x+3) => 6x -17 = 4x +6 => 2x = 23 =>x=23/2
then just plug x into LB and AB
5 0
3 years ago
Find the value of x for which the lines a and b are parallel. answer : 23 20 2 19
Afina-wow [57]

Answer:

x=19

Step-by-step explanation:

These two angles are alternate interior angles and alternate interior angles are equal when the lines are parallel

148 = 7x+15

Subtract 15 from each side

148-15 =7x+15-15

133 = 7x

Divide by 7

133/7 = 7x/7

19 =x

4 0
3 years ago
Condense each expression to a single logarithm <br>36 ln x + 6 ln y
Igoryamba

Answer:

36x+6y

Step-by-step explanation:


8 0
3 years ago
Read 2 more answers
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