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3 years ago

The attached screenshot is the question. The options are yes or no for each question

Mathematics

1 answer:

lions [1.4K]3 years ago

4 0

Answer:

No, yes, yes

Step-by-step explanation:

All of those question the rules of irrational vs rational numbers.

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6q+4-q+5<br><br> what is the answer pls tell me

JulsSmile [24]

Answer:

5q+9

Step-by-step explanation:

6q minus a is 5 q so then it would equal 5q+9

which is 4+5 equals 9

6 0

3 years ago

Read 2 more answers

What is the equation of a line that has a y-intercept of 4 and an x-intercept of -2?

Furkat [3]

Answer:

y = 2x + 4

Step-by-step explanation:

Two points on this line are (-2, 0) and (0, 4). Going from the first to the second, x increases by 2 (this is the 'run') and y increases by 4 ('rise').

Thus, the slope of this line is m = rise / run = 4/2 = 2

Using the slope-intercept formula, we get y = mx + b = 2x + b

Let x = -2 and y = 0 to find b: 0 = 2(-2) + b, so b = 4, and the desired equation is then:

y = 2x + 4

6 0

3 years ago

Which equation represents the circle described? (in picture)

Sedaia [141]

You are right. In circle equation we have =r² so the correct answer will be C od D. But we have to calculate the center of a circle.

$x^2+y^2-8x-6y+24=0\quad|-24\\\\(x^2-8x)+(y^2-6y)=-24$

Complete the square:

$x^2-8x+a^2\implies a=\dfrac{8x}{2x}=4\implies x^2-8x+4^2\\\\\\y^2-6y+b^2\implies b=\dfrac{6y}{2y}=3\implies y^2-6y+3^2$

And we have:

$(x^2-8x)+(y^2-6y)=-24\\\\(x^2-8x+4^2)-4^2+(y^2-6y+3^2)-3^2=-24\\\\
(x-4)^2+(y-3)^2-16-9=-24\\\\(x-4)^2+(y-3)^2-25=-24\quad|+25\\\\\boxed{(x-4)^2+(y-3)^2=1}$

So the answer is C (the same left side of equation).

6 0

3 years ago

Which statements are true about the expression

bazaltina [42]

Answer:

Wheres the question?

6 0

3 years ago

Find the radius of the circle with equation x²+y²+8x+8y+28=0

galina1969 [7]

<h2>Hello!</h2>

The answer is:

Center: (-4,-4)

Radius: 2 units.

To solve the problem, using the given formula of a circle, we need to find its standard equation form which is equal to:

$(x-h)^{2}+(y-k)^{2}=r^{2}$

Where:

"h" and "k" are the coordinates of the center of the circle and "r" is its radius.

So, we need to complete the square for both variable "x" and "y".

The given equation is:

$x^{2}+y^{2}+8x+8y+28=0$

So, solving we have:

$x^{2}+y^{2}+8x+8y=-28$

$(x^2+8x+(\frac{8}{2})^{2})+(y^2+8y+(\frac{8}{2})^{2})=-28+((\frac{8}{2})^{2})++(\frac{8}{2})^{2})\\\\(x^2+8x+16 )+(y^2+8y+16)=-28+16+16\\\\(x^2+4)+(y^2+4)=4$

$(x^2-(-4))+(y^2-(-4))=4$

Now, we have that:

$h=-4\\k=-4\\r=\sqrt{4}=2$

So,

Center: (-4,-4)

Radius: 2 units.

Have a nice day!

Note: I have attached a picture for better understanding.

3 0

3 years ago