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3 years ago

13

Choose the equations that are not linear equations.

1 answer:

jeyben [28]3 years ago

5 0

Answer: x/2-y^3=1 & x^2+y=4

Step-by-step explanation:

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Find the next two terms of the sequence. 2, 6, 10, 14, ... (1 point)

Naddik [55]

The sequence given is 2, 6, 10, 14
Noticing this sequence, we will find that the sequence increases by 4 each time. This means that:
The next number = the current number + 4

In other words:
2
2+4 = 6
6+4 = 10
10+4 = 14
14+4 = 18

Therefore, the next number in the sequence would be 18

7 0

3 years ago

Read 2 more answers

How do you simplify this?<br>x²y+xy² / y²+2/5 × xy

polet [3.4K]

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

How do you simplify this?
x²y+xy² / y²+2/5 × xy

$\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}$

$\sf\frac{ { x }^{ 2 } y+x { y }^{ 2 } }{ { y }^{ 2 } + \frac{ 2 }{ 5 } \times xy } \\$

Factor the expressions that are not already factored.

_____

<u>How </u><u>to</u><u> factorise</u><u> </u><u>:</u><u>-</u>

<u>NUMERATOR</u> $\downarrow$

$\sf \: x ^ { 2 } y + x y ^ { 2 }$

Factor out xy.

$\sf \: xy\left(x+y\right)$

<u>DENOMINATOR</u> $\downarrow$

$\sf{ y }^{ 2 } + \frac{ 2 }{ 5 } \times xy \\$

Factor out 1/5.

$\sf \: {\frac{1}{5}y\left(2x+5y\right)} \\$

_____

Continuing...

$\sf\frac{xy\left(x+y\right)}{\frac{1}{5}y\left(2x+5y\right)} \\$

Cancel out y in both the numerator and denominator.

$\sf\frac{x\left(x+y\right)}{\frac{1}{5}\left(2x+5y\right)} \\$

Expand the expression.

$\sf\frac{x^{2}+xy}{\frac{2}{5}x+y} \\$

This can further simplified to as $\downarrow$

$= \boxed{\boxed{\bf\frac{5x\left(x +y\right)}{2x+5y}}}$

3 0

2 years ago

A puzzle is shown below. what is the area of the shaded portions of the puzzle?

kolezko [41]

Answer:

it's a parallelogram

Step-by-step explanation:

May ot helpful

5 0

2 years ago

Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.

lorasvet [3.4K]

Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

$PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ$

Likewise:

$\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ$

Since they all measure 60°.

Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

$m\angle PAC=m\angle PAB+m\angle BAC$

Likewise:

$m\angle QAB=m\angle QAC+m\angle BAC$

Since ∠QAC ≅ ∠PAB:

$m\angle PAC=m\angle QAC+m\angle BAC$

And by substitution:

$m\angle PAC=m\angle QAB$

Thus:

$\angle PAC\cong \angle QAB$

Then by SAS Congruence:

$\Delta PAC\cong \Delta BAQ$

And by CPCTC:

$PC\cong BQ$

5 0

2 years ago

Read 2 more answers

To graph the line y = 3x -2, I should *

ipn [44]

Answer:

The answer is D because you should starts at -2 on the y-axis and moves up 3 and right 1.

4 0

3 years ago