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3 years ago

12

Two triangles are similar if there is a correspondence between their angles and their sides so that all corresponding angles are

______ and all corresponding sides are _______.

1 answer:

Setler79 [48]3 years ago

4 0

Answer:

see the explanation

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional (this ratio is called the scale factor) and its corresponding angles are congruent

therefore

Two triangles are similar if there is a correspondence between their angles and their sides so that all corresponding angles are <u><em>congruent </em></u> and all corresponding sides are <u><em>proportional</em></u>

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Kay [80]

The mistake is that that is an equilateral triangle with the same answer for each side which is 6 so AC=6 BC=6 BA=6 all sides equal 6

3 0

3 years ago

What are the solutions of the equation 9x4 – 2x2 – 7 = 0? Use u substitution to solve. tions of the equation 9

skelet666 [1.2K]

Answer:

Step-by-step explanation:

Let $u^2=x^4\\u = x^2$

Subbing in:

$9u^2-2u-7=0$

a = 9, b = -2, c = -7

The product of a and c is the aboslute value of -63, so a*c = 63. We need 2 factors of 63 that will add to give us -2. The factors of 63 are {1, 63}, (3, 21}, {7, 9}. It looks like the combination of -9 and +7 will work because -9 + 7 = -2. Plug in accordingly:

$9u^2-9u+7u-7=0$

Group together in groups of 2:

$(9u^2-9u)+(7u-7)=0$

Now factor out what's common within each set of parenthesis:

$9u(u-1)+7(u-1)=0$

We know this combination "works" because the terms inside the parenthesis are identical. We can now factor those out and what's left goes together in another set of parenthesis:

$(u-1)(9u+7)=0$

Remember that $u=x^2$

so we sub back in and continue to factor. This was originally a fourth degree polynomial; that means we have 4 solutions.

$(x^2-1)(9x^2+7)=0$

The first two solutions are found withing the first set of parenthesis and the second two are found in other set of parenthesis. Factoring $(x^2-1)$ gives us that x = 1 and -1. The other set is a bit more tricky. If

$9x^2+7=0$ then

$9x^2=-7$ and

$x^2=-\frac{7}{9}$

You cannot take the square root of a negative number without allowing for the imaginary component, i, so we do that:

$x=$ ± $\sqrt{-\frac{7}{9} }$

which will simplify down to

$x=$ ± $\frac{\sqrt{7} }{3}i$

Those are the 4 solutions to the quartic equation.

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3 years ago

A carpenter has a board that is 10 feet long. He wants to make 6 table legs that are all the same length. What is the longest ea

Zielflug [23.3K]

1 board that is 10 feet long and make that into 6 table legs. Take the amount we have over the amount we want, so 10 over 6 which is 1.6666667 feet which is the longest each leg can be for a 10-foot board.

3 0

3 years ago

Read 2 more answers

The 65 students in a classical music lecture class were polled, with the following results: 37 like Wolfgang Amadeus Mozart 36 l

Anna35 [415]

Answer:

a) 25

b) 30

c) 10

d) Not Mozart, 6

e) 2

Step-by-step explanation:

We use a Venn Diagram to solve this question.

I am going to say that:

A are the students who like Mozart.

B are the students who like Beethoven

C are the students who like Haydn.

We have that:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which a are those who only like Mozart, $(A \cap B)$ are those who like Mozart and Beethoven, $(A \cap C)$ are those who like Mozart and Haydn and $(A \cap B \cap C)$ are those who like all three of them.

By the same logic, we have that:

$B = b + (A \cap B) + (B \cap C) + (A \cap B \cap C)$

$C = c + (B \cap C) + (A \cap C) + (A \cap B \cap C)$

We start finding these values from the intersection:

8 like all three composers

This means that $A \cap B \cap C = 8$

14 like Beethoven and Haydn

This means that:

$(B \cap C) + (A \cap B \cap C) = 14$

So

$B \cap C = 6$

21 like Mozart and Haydn

This means that:

$(A \cap C) + (A \cap B \cap C) = 21$

Then

$A \cap C = 13$

14 like Mozart and Beethoven

This means that:

$(A \cap B) + (A \cap B \cap C) = 14$

$A \cap B = 6$

31 like Franz Joseph Haydn

This means that C = 31. So

$C = c + (B \cap C) + (A \cap C) + (A \cap B \cap C)$

$31 = c + 6 + 13 + 8$

$c = 4$

36 like Ludwig van Beethoven

This means that $B = 36$

So

$B = b + (A \cap B) + (B \cap C) + (A \cap B \cap C)$

$36 = b + 6 + 6 + 8$

$b = 16$

37 like Wolfgang Amadeus Mozart

This means that A = 37. Then

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

$37 = a + 6 + 13 + 8$

$a = 10$

a. exactly two of these composers?

$(A \cap B) + (A \cap C) + (B \cap C) = 6 + 13 + 6 = 25$

b. exactly one of these composers?

$a + b + c = 10 + 16 + 4 = 30$

c. like only Mozart?

$a = 10$

d. like Beethoven and Haydn, but not Beethoven?

I will use not Mozart.

So $B \cap C = 6$

Not Mozart, 6.

e. like none of these composers?

At least 1:

$(A \cup B \cup C) = a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 10 + 16 + 4 + 6 + 13 + 6 + 8 = 63$

The total is 65

So 65 - 63 = 2 like none of these composers

4 0

3 years ago

Use the properties of exponents to solve for x in the following:

wariber [46]

Hello,
Please, see the attached file.
Thanks.

3 0

3 years ago