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2 years ago

HELP ASAP!!! select all the descriptions that apply to the transformation!

Mathematics

1 answer:

maxonik [38]2 years ago

6 0

Answer: A and B

Step-by-step explanation:

The shape is moving from right to left and down

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Ulleksa [173]

Step-by-step explanation:

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1 year ago

If two zeroes of a polynomial x3+5x2+7x+3 are -1 and-3 then find the third Zero

belka [17]

Answer:

Third zero is -1.

Step-by-step explanation:

consider the given polynomial is

$P(x)=x^3+5x^2+7x+3$

It is given that -1 and -3 are two zeroes, therefore (x+1) and (x+3) are two factors of given polynomial.

The given polynomial can be rewritten as

$P(x)=x^3+x^2+4x^2+4x+3x+3$

$P(x)=x^2(x+1)+4x(x+1)+3(x+1)$

$P(x)=(x+1)(x^2+4x+3)$

Now, splitting the middle term, we get

$P(x)=(x+1)(x^2+3x+x+3)$

$P(x)=(x+1)(x(x+3)+1(x+3))$

$P(x)=(x+1)(x+1)(x+3)$

$P(x)=(x+1)^2(x+3)$

Here, power of (x+1) is 2. It means the multiplicity of zero -1 is 2.

So, three zeroes are -1, -1 and -3.

Therefore the third zero is -1.

7 0

2 years ago

When was geometric sequence discovered and who made it

lord [1]

Geometric sequences are mostly found in Book IX of Elements by Euclid in 300 B.C. Euclid of Alexandria, a Greek mathematician also considered the "Father of Geometry" was the main contributor of this theory. Geometric sequences and series are one of the easiest examples of infinite series with finite sums. Geometric sequences and series have played an important role in the early development of calculus, and have continued to be a main case of study in convergence of series. Geometric sequences and series are used a lot in mathematics, and they are very important in physics, engineering, biology, economics, computer science, queuing theory, and finance.<span> It was included in Euclid's book </span>Elements<span> that was part of a composition of other math theories for people that became very popular because it was the first collection that showed alot of the main math theories together featured simply.</span>

6 0

2 years ago

How do I get (tan^2(x)-sin^2(x))/tan(x) equal to (sin^2(x))/cot(x)

shepuryov [24]

$LHS\\ \\ =\frac { \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } }{ \tan { x } } \\ \\ =\frac { 1 }{ \tan { x } } \left( \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } \right)$

$\\ \\ =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^{ 2 }{ x } }{ \cos ^{ 2 }{ x } } -\frac { \sin ^{ 2 }{ x\cos ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } \right) \\ \\ =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^{ 2 }{ x-\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } } }{ \cos ^{ 2 }{ x } } \right)$

$\\ \\ =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^{ 2 }{ x\left( 1-\cos ^{ 2 }{ x } \right) } }{ \cos ^{ 2 }{ x } } \\ \\ =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^{ 2 }{ x\cdot \sin ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } \\ \\ =\frac { \cos { x } \sin ^{ 4 }{ x } }{ \sin { x\cos ^{ 2 }{ x } } } \\ \\ =\frac { \sin ^{ 3 }{ x } }{ \cos { x } }$

$\\ \\ =\sin ^{ 2 }{ x } \cdot \frac { \sin { x } }{ \cos { x } } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \frac { \cos { x } }{ \sin { x } } } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \cot { x } } \\ \\ =\frac { \sin ^{ 2 }{ x } }{ \cot { x } } \\ \\ =RHS$

8 0

2 years ago

Simplified by combing like terms 4z-5/6z

zhannawk [14.2K]

Answer:

19z/6

Step-by-step explanation:

4 0

2 years ago