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3 years ago

I don’t get this help

Mathematics

2 answers:

Ahat [919]3 years ago

8 0

Answer:

its the 4th one

Step-by-step explanation:

givi [52]3 years ago

8 0

The answer is the last one 4th

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I need help with this one question.

Tom [10]

Answer:

answer a

Step-by-step explanation:

3 0

4 years ago

Given right triangle ABC with altitude BD drawn to hypotenuse AC. If AC =16 and DC=5 what is the length of BC in the simplest ra

Nana76 [90]

The length of BC is $4 \sqrt{5}$ .

Solution:

Given ABC is a right triangle.

AC is the hypotenuse and BD is the altitude.

AB and BC are legs of the triangle ABC.

AC = 16 and DC = 5

<u>Leg rule of geometric mean theorem:</u>

$\frac{\text { hypotenuse }}{\text { leg }}=\frac{\text { leg }}{\text { part }}$

$\Rightarrow \frac{AC}{BC}=\frac{BC}{DC}$

$\Rightarrow \frac{16}{x}=\frac{x}{5}$

Do cross multiplication.

$\Rightarrow 16\times 5 = x\times x$

$\Rightarrow 80= x^2$

$\Rightarrow 16\times 5= x^2$

Taking square root on both sides.

$\Rightarrow \sqrt{16\times 5} = \sqrt{x^2}$

$\Rightarrow \sqrt{4^2\times 5} = \sqrt{x^2}$

square and square roots get canceled, we get

$\Rightarrow 4\sqrt{ 5} = x$

The length of BC is $4 \sqrt{5}$ .

7 0

3 years ago

Which of the sequences is an arithmetic sequence?

Dennis_Churaev [7]

Answer:

B. -3, -10, -17, -24, -31

Step-by-step explanation:

An Arithmetic sequence is defined as a set of numbers with a common difference. In answer choice B, it is seen that the difference between each number is seven.

A is incorrect as one to eight has a difference of seven, while the other numbers have a difference of eight.

C is incorrect as it skips the number 12 between 9 and 15, thus breaking the sequence.

D is incorrect as the numbers jump from positive to negative, thus not having a common difference.

Hope this helps!

7 0

4 years ago

Read 2 more answers

Which products result in a perfect square trinomial? Select three. (-x+9)(-x-9), (xy+x)(xy+x), (2x-3)(-3+2x), (16-x^2)(x^2-16),

GrogVix [38]

Answer:

(2x-3)(-3+2x),(16-x^2)(x²-16),4y^2+25)(25+4y^2)

Step-by-step explanation:

I know that the rest of the products didn't make a perfect square cause they don´t aren´t squared,I also know that they are the odd ones out among the rest of the product so yeah,I am not too sure if I got it right...sorry if I didn´t.

3 0

4 years ago

Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:

professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is $\{1,2,3,4\}$ .

a) We need to find a relation R reflexive and transitive that contain the relation $R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}$

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

1R4 and 4R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
4R1 and 1R2, then 4 must be related with 2.

Therefore $\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\}$ is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

since 1R2, then 2 must be related with 1.
since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

1R2 and 2R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
2R1 and 1R4, then 2 must be related with 4.
4R1 and 1R2, then 4 must be related with 2.
2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

1 must be related with 1,

2 must be related with 2,

3 must be related with 3,

4 must be related with 4

For be symmetric

since 1R2, 2 must be related with 1,
since 1R4, 4 must be related with 1.

For be transitive

Since 4R1 and 1R2, 4 must be related with 2,
since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

5 0

3 years ago