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3 years ago

Do you know the answer??????

Mathematics

2 answers:

Daniel [21]3 years ago

7 0

It’s not b it’s A. No y values repeat

sveticcg [70]3 years ago

4 0

Answer:

Step-by-step explanation:

because when you divide all the answers their hours of training its all different answers

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PERGUNTA 1

Reptile [31]

C I guess

I don’t understand

5 0

3 years ago

HELP IS NEED QUICKLY I WILL GIVE BRAINLIEST

Alex Ar [27]

Answer:

Well add the first numbers and get 10.34 you can add up to at least 5 to 6 toppings.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

(20 points)Solve the equation cos x = sin (x + 22)

Law Incorporation [45]

Answer:

$\bold{x = 34^\circ}$

Step-by-step explanation:

Given:

$cos x = sin (x + 22^\circ)$

To solve:

The given equation.

Solution:

First of all, let us consider an important property of sine and cosine.

$sin(90^\circ-\theta )=cos\theta$

$cos(90^\circ-\theta )=sin\theta$

We can apply above property to solve for $x$ as per given equation.

$cos x = sin (x + 22^\circ)$

Changing $cosx$ to sine form:

$cosx=sin(90^\circ-x)$

$cosx=sin(90^\circ-x) = sin(x+22^\circ)$

$\therefore 90^\circ-x=x+22^\circ\\\Rightarrow 90^\circ-22^\circ=x+x\\\Rightarrow 2x=68^\circ\\\Rightarrow \bold{x = 34^\circ}$

So, solution to the equation $cos x = sin (x + 22^\circ)$ is:

$\bold{x = 34^\circ}$

7 0

3 years ago

What is the difference of the polynomials? (–2x3y2 + 4x2y3 – 3xy4) – (6x4y – 5x2y3 – y5)

ANEK [815]

For the answer to the question above, I'll provide my solutions to my answers for the problem below.

(–2x3y2 + 4x2y3 – 3xy4) – (6x4y – 5x2y3 – y5)

(−2x3)(y2)+4x2y3+−3xy4+−1(6x4y)+−1(−5x2y3)+−1(−y5)

(−2x3)(y2)+4x2y3+−3xy4+−6x4y+5x2y3+y5

−2x3y2+4x2y3+−3xy4+−6x4y+5x2y3+y5

−2x3y2+4x2y3+−3xy4+−6x4y+5x2y3+y5

(−6x4y)+(−2x3y2)+(4x2y3+5x2y3)+(−3xy4)+(y5)

−6x4y+−2x3y2+9x2y3+−3xy4+y5
So the answer is,
= −6x4y−2x3y2+9x2y3−3xy4+y5

I hope this helps

4 0

3 years ago

Read 2 more answers

Simplify each expression. Use positive exponents. A^4b^-3 ab^-2

choli [55]

$\dfrac{a^4b^{-3}}{ab^{-2}}$

$= {a^{4-1}b^{-3+2}}$

$= {a^{3}b^{-1}}$

$= \dfrac{a^3}{b}$

6 0

2 years ago

Read 2 more answers