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2 years ago

If x > 0 and y > 0, where is the point (x, y) located?

Mathematics

1 answer:

Scrat [10]2 years ago

8 0

Given, x > 0, y > 0.
So, the values of x and y are less than 0, that means, they are negative integers.
If we divide a Cartesian plane with the x-axis and y-axis, we get four quadrants.
1st Quadrant : (+,+)
2nd Quadrant : (-,+)
3rd Quadrant : (-,-)
4th Quadrant : (+,-)
Since the values of x and y are both negative, so (x, y) is located in the <em><u>3rd Quadrant</u></em>.

Hope you could get an idea from here.

Doubt clarification - use comment section

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Volume of the cylinder with diameter of 8mm and height of 8mm

alexdok [17]

Answer:

V = 402.12385mm

Step-by-step explanation:

When the diameter is 8mm, the radius is 4mm.

V = (π(4^2))8

V = (16π)8

V = (50.2654)8

V = 402.12385

V ≈ 402, 402.1, 402.12, 402.124

4 0

2 years ago

Help, I have a time limit for this

salantis [7]

Answer:

I believe that it is the first one.

Step-by-step explanation:

5 0

2 years ago

1. D=5.9 ft.

Shtirlitz [24]

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

Here's the solution ~

As we know, we can calculate the circumference of a circle in terms of its diameter as :

$\qquad \sf \dashrightarrow \:c = \pi d$

where, c = circumference and d = diameter

And also, circumference of circle is terms of radius (r) is :

$\qquad \sf \dashrightarrow \:c = 2\pi r$

Now, let's move on to questions ~

<h3>First </h3>

$\qquad \sf \dashrightarrow \:3.14 \times 5.9$

$\qquad \sf \dashrightarrow \: \approx18.53 \: ft$

<h3 />

・．━━━━━━━†━━━━━━━━━．・

<h3>Second</h3><h3 /><h3 /><h3 /><h3> $\qquad \sf \dashrightarrow \:3.14 \times 3.2$ </h3>

$\qquad \sf \dashrightarrow \: \approx10.048 \: ft$

・．━━━━━━━†━━━━━━━━━．・

<h3>Third</h3>

$\qquad \sf \dashrightarrow \:3.14 \times 6.1$

$\qquad \sf \dashrightarrow \: \approx19.15 \: ft$

・．━━━━━━━†━━━━━━━━━．・

<h3>Fourth</h3>

$\qquad \sf \dashrightarrow \:2×3.14 \times 3.7$

$\qquad \sf \dashrightarrow \: \approx2×11.62 \: m$

$\qquad \sf \dashrightarrow \: \approx23.24 \: m$

・．━━━━━━━†━━━━━━━━━．・

<h3>Fifth </h3>

$\qquad \sf \dashrightarrow \:2×3.14 \times 6.2$

$\qquad \sf \dashrightarrow \: \approx2× 19.47 \: units$

$\qquad \sf \dashrightarrow \: \approx38.94 \: m$

・．━━━━━━━†━━━━━━━━━．・

<h3>Sixth</h3>

$\qquad \sf \dashrightarrow \:2×3.14 \times 5.1$

$\qquad \sf \dashrightarrow \: \approx2×16.01 \: ft$

$\qquad \sf \dashrightarrow \: \approx \: 32.02m$

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

7 0

2 years ago

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

2 years ago

In a school band, the ratio of student late who play the tuba to the students who play the trumpets is 1:6. The ratio of student

AlexFokin [52]

A lot, 1:6 + 3:4 +3 = .116789465 hope this helps....

7 0

3 years ago