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Gennadij [26K]
3 years ago
13

What else would need to be congruent to show that ABC = XYZ by ASA?

Mathematics
1 answer:
Rom4ik [11]3 years ago
8 0

Answer:

Option B AC≅XZ

Step-by-step explanation:

we know that

ASA (angle, side, angle) means that we have two congruent triangles where we know two angles and the included side are equal

In this problem we have

∠X≅∠A

∠Z≅∠C

In the triangle ABC the included side between the angles ∠A and ∠C is the side AC

and

In the triangle XYZ the included side between the angles ∠X and ∠Z is the side XZ

therefore

AC≅XZ

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Which of the following is a trinomial?
Sedaia [141]

Answer:

A) c^2 + c + 6

Step-by-step explanation:

We have been given four choices out of which we need to select the choice which is trinomial.

Trinomial means polynomial having three terms.

A) c2 + c + 6

It has 3 terms.

B) c2 − 16

It has 2 terms.

C) −8c

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D) c3 + 4c2 − 12c + 7

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We see that only choice A) has three terms.

Hence correct choice is A) c^2 + c + 6


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3 years ago
An angle measures 67.2° less than the measure of a supplementary angle. What is the measure of each angle?
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Two supplementary angles always adds up to 180 degrees. If an angle is 67.2 degrees less than the measure of a supplementary angle, an equation can be formed where "A" represents the larger angle, and (A - 67.2) represents the smaller angle.

(A - 67.2)+A=180

(Combine the A's)

2A - 67.2=180

(Add 67.2 to both sides)

2A = 247.2

(Divide by 2 to both sides)

A = 123.6

123.6 degrees is the larger angle.

The smaller angle is represented by (A - 67.2). We now know the value of A so lets plug it in. (123.6 - 67.2) = 56.4. 56.4 is the value of the smaller angle.

We can double check to make sure we got the right answer.

123.6 + 56.4 = 180
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Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it
Zanzabum

Answer:

\beta=45\degree\:\:or\:\:\beta=135\degree

Step-by-step explanation:

We want to solve \tan \beta \sec \beta=\sqrt{2}, where 0\le \beta \le360\degree.

We rewrite in terms of sine and cosine.

\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}

\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}

Use the Pythagorean identity: \cos^2\beta=1-\sin^2\beta.

\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}

\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)

\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta

\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0

This is a quadratic equation in \sin \beta.

By the quadratic formula, we have:

\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }

\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }

\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }

\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }

\sin \beta=\frac{2}{2\cdot \sqrt{2} } or \sin \beta=\frac{-4}{2\cdot \sqrt{2} }

\sin \beta=\frac{1}{\sqrt{2} } or \sin \beta=-\frac{2}{\sqrt{2} }

\sin \beta=\frac{\sqrt{2}}{2} or \sin \beta=-\sqrt{2}

When \sin \beta=\frac{\sqrt{2}}{2} , \beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )

\implies \beta=45\degree\:\:or\:\:\beta=135\degree on the interval 0\le \beta \le360\degree.

When  \sin \beta=-\sqrt{2}, \beta is not defined because -1\le \sin \beta \le1

4 0
3 years ago
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