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3 years ago

please help me anyone ?!!!!!

Mathematics

1 answer:

NeTakaya3 years ago

4 0

The answer to this is B

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Im not sure about this im confused

Maru [420]

Answer:

Step-by-step explanation:

8 0

3 years ago

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Solve the following simultaneous equations : with the help of steps 1. x + 2y = 1, 3x - y = 17

IRISSAK [1]

Answer:

x = 5, y = -2

Step-by-step explanation:

Given equations:

x + 2y = 1
3x - y = 17

From the first equation we get:

x = 1 -2y

Substitute x in the second equation, find y:

3x - y = 17
3*(1-2y) - y = 17
3 - 6y - y = 17
-7y = 17 - 3
-7y = 14
y = -14/7
y = -2

Find x:

x = 1 - 2y
x = 1 - 2*(-2)
x = 1 + 4
x = 5

6 0

3 years ago

Helppp plzzz right nowwwwwwww

Pavlova-9 [17]

-4/3

Explanation: All parallel lines have the same slope

7 0

3 years ago

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The polynomial x^2+3x-1 is a factor of x^4+3x^3-2x^2-3x+1 true or false?

pashok25 [27]

Answer:

Yes, it is true that $x^2+3x-1$ is a factor of $x^4+3x^3-2x^2-3x+1$ .

Step-by-step explanation:

Let us try to factorize $x^4+3x^3-2x^2-3x+1$

$x^4+3x^3-2x^2-3x+1\\\Rightarrow x^4-2x^2+1-3x+3x^3$

Let us try to make a whole square of the given terms:

$\Rightarrow (x^2)^2-2\times x^2 \times 1+1^2+3x^3-3x\\\Rightarrow (x^2-1)^2+3x^3-3x\\$

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Formula used above:

$a^{2} -2 \times a \times b +b^{2} = (a-b)^2$

In the above equation, we had $a = x, b = 1$ .

--------------

Further solving the above equation, taking $3x$ common out of $3x^3-3x$

$\Rightarrow (x^2-1)^2+3x(x^2-1)\\$

Taking $(x^{2} -1)$ common out of the above term:

$\Rightarrow (x^2-1)((x^2-1)+3x)\\\Rightarrow (x^2-1)(x^2+3x-1)$

So, the two factors are $(x^2-1)\ and\ (x^2+3x-1)$ .

$\therefore$ The statement that $x^2+3x-1$ is a factor of $x^4+3x^3-2x^2-3x+1$ is True.

7 0

3 years ago

**TIMED**

azamat

Coterminal angles are found by adding/subtracting 2 $\pi$ or 360, so D.

6 0

3 years ago

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￼please help me anyone ?!!!!!

please help me anyone ?!!!!!