Answer:
The coordinates of the point that is a reflection of Y(-4, -2) across the x-axis are (
-4,2).
The coordinates of the point that is a reflection of Y across the y-axis are (
4,-2).
Step-by-step explanation:
<em>Reflection across x-axis</em>
<em>The rule used for Reflection across x-axis is that y-coordinate becomes negated while x coordinate remains same.</em>
So,
The coordinates of the point that is a reflection of Y(-4, -2) across the x-axis are (
-4,2).
Because according to definition, x-coordinate remains same, while y-coordinate is negated. So x-coordinate = -4, y-coordinate = 2
<em>Reflection across y-axis</em>
<em>The rule used for Reflection across y-axis is that x-coordinate becomes negated while y coordinate remains same.</em>
So,
The coordinates of the point that is a reflection of Y across the y-axis are (
4,-2).
Because according to definition, y-coordinate remains same, while x-coordinate is negated. So x-coordinate = 4, y-coordinate = -2
Answer: 44
Step-by-step explanation:
we will find RN and NQ, then add together to give us RQ.
To find RN;
RP= 17 PN = 15 and RN =?
using pythagoras theorem,
adj^2 = hyp^2 - opp^2
RN^2 = RP^2 - PN^2
?^2 = 17^2 - 15^2
?^2 = 17^2 - 15^2
?^2 = 289 - 225
?^2 = 64
? = √64
? = 8
RN=8
To find NQ,
PN = 15 PQ=39 and NQ=?
using pythagoras theorem
NQ^2 = PQ^2 - PN^2
?^2 = 39^2 - 15^2
?^2 = 1521 - 225
?^2 = 1296
? = √1296
? = 36
NQ= 36
RQ = RN + NQ
RQ= 8 + 36
RQ=44
Answer:
B
Step-by-step explanation:
Also the answer must be under 9 so that rules out option D
Because We know CE = 2 and AD is visibly larger than CE, 6We can rule out option A
And that brings us down to B and C
CB is equal to 6, And I'm no genius, but I'm pretty sure AD is DEFINATLY not equal to 6.
so the answer is B
Answer:
Step-by-step explanation:
[in picture]
Answer:
-937.5π
Step-by-step explanation:
F (r) = r = (x, y, z) the surface equation z = 3(x^2 + y^2) z_x = 6x, z_y = 6y the normal vector n = (- z_x, - z_y, 1) = (- 6x, - 6y, 1)
Thus, flux ∫∫s F · dS is given as;
∫∫ <x, y, z> · <-z_x, -z_y, 1> dA
=∫∫ <x, y, 3x² + 3y²> · <-6x, -6y, 1>dA , since z = 3x² + 3y²
Thus, flux is;
= ∫∫ -3(x² + y²) dA.
Since the region of integration is bounded by x² + y² = 25, let's convert to polar coordinates as follows:
∫(θ = 0 to 2π) ∫(r = 0 to 5) -3r² (r·dr·dθ)
= 2π ∫(r = 0 to 5) -3r³ dr
= -(6/4)πr^4 {for r = 0 to 5}
= -(6/4)5⁴π - (6/4)0⁴π
= -937.5π
