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1 year ago

Select the best answer for the question.

Mathematics

1 answer:

kherson [118]1 year ago

8 0

Answer:

A. 1/2 and 3/2

Step-by-step explanation:

You want to identify like fractions among the sets {1/2, 3/2}, {2/1, 2/3}, {10/10, 5/5}, and {7/4, 4/7}.

<h3>Like fractions</h3>

Like fractions are fractions that have the <em>same denominator</em>. Among the sets offered, the only one with denominators the same is ...

{1/2, 3/2}

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Serggg [28]

Answer:

value of sides YZ is 36 cm

Step-by-step explanation:

Similar triangles states that the length of the corresponding sides are in proportion.

Given that: ΔABC is similar to ΔXYZ

then;

Corresponding sides are in proportion i.e

$\frac{AB}{XY}=\frac{BC}{YZ}=\frac{AC}{XZ}$ .....[1]

As per the statement:

side AB = 6 cm, side BC =18 cm and side XY = 12 cm.

Substitute these in [1] to solve for side YZ;

$\frac{6}{12}= \frac{18}{YZ}$

$\frac{1}{2} = \frac{18}{YZ}$

By cross multiply we have;

$YZ = 36$ cm

Therefore, the value of sides YZ is 36 cm

4 0

2 years ago

Read 2 more answers

Suppose a triangle has sides a, b, and c, and the angle opposite the side of length b is obtuse. What must be true?

anyanavicka [17]

Answer: D. $a^2 + b^2 < c^2$

Step-by-step explanation:

In the given figure, we have an obtuse triangle which has sides a, b, and c, and the angle opposite the side of length b is obtuse.

∵ In a triangle, the side opposite to the largest angle is largest.

Thus, the largest side in then given obtuse triangle= c

The Pythagorean inequalities theorem says that If a triangle is obtuse than the square of the largest side is greater than the sum of square of other two sides.

Therefore, we have

$a^2 + b^2 < c^2$

5 0

2 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

Jase used a drawing program on the computer at school to create the figure below using a

Xelga [282]

Answer : 15.75 square inches

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

I am giving away 10 points to anybody who provides correct answers with explanations.

QveST [7]

Answer: