Can you guys help and plz be quick
1 answer:
Answer:
Step-by-step explanation:
Use the zero-interval test [test point (0, 0)] to verify them as false or true, and to determine whether we shade the shared opposite portions [a shared portion that does NOT contain the origin] or the shared portions that DO contain the origin:
So, this system will be shaded in a shared opposite portion.
*NOTE; Both inequalities have an identical y-intercept of so both inequalities must intersect the y-axis at
and since there are already lines labeled as <em>dashed</em><em> </em>and solid, we do not have to worry about it in this case.
** None of the other graphs match this, so the bottom right graph has to be it, considering it cut off.
I am joyous to assist you anytime.
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To find out if a point is inside, on, or outside a circle, we need to substitute the ordered pair into the equation of the circle:
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.
If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle. If the left-hand side is equal to r^2, the point is ON the circle.
Finally, if the left-hand side is greater than r^2, the point is OUTSIDE the circle.
For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
(x-xc)^2+(y-yc)^2=r^2 => (x-3)^2+y^2=7^2
A. (-1,1),
(x-3)^2+y^2=7^2 => (-1-3)^2+1^2=16+1=17 <49 [inside circle]
B. (10,0)
(x-3)^2+y^2=7^2 => (10-3)^2+0^2=49+0=49 [on circle]
C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49 [outside circle]
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.
If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle. If the left-hand side is equal to r^2, the point is ON the circle.
Finally, if the left-hand side is greater than r^2, the point is OUTSIDE the circle.
For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
(x-xc)^2+(y-yc)^2=r^2 => (x-3)^2+y^2=7^2
A. (-1,1),
(x-3)^2+y^2=7^2 => (-1-3)^2+1^2=16+1=17 <49 [inside circle]
B. (10,0)
(x-3)^2+y^2=7^2 => (10-3)^2+0^2=49+0=49 [on circle]
C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49 [outside circle]
See the attached figure.
<span>ad is a diameter of the circle with center p
</span>
∵ pd = radius = 7 ⇒⇒⇒ ∴ ad = 2 * radius = 2 * 7 = 14
∵ ae = 4 ⇒⇒⇒ ∴ ed = ad - ae = 14 - 4 = 10
∵ ad is a diameter
Δ acd is a triangle drawn in a half circle
∴ Δ acd is a right triangle at c
∵ bc ⊥ ad at point e
By applying euclid's theorem inside Δ acd
∴ ce² = ae * ed
∴ ce² = 4 * 10 = 40
∴ ce = √40 = 2√10 ≈ 6.325
<span>ad is a diameter of the circle with center p
</span>
∵ pd = radius = 7 ⇒⇒⇒ ∴ ad = 2 * radius = 2 * 7 = 14
∵ ae = 4 ⇒⇒⇒ ∴ ed = ad - ae = 14 - 4 = 10
∵ ad is a diameter
Δ acd is a triangle drawn in a half circle
∴ Δ acd is a right triangle at c
∵ bc ⊥ ad at point e
By applying euclid's theorem inside Δ acd
∴ ce² = ae * ed
∴ ce² = 4 * 10 = 40
∴ ce = √40 = 2√10 ≈ 6.325