All categories

2 years ago

Can someone please solve this

Mathematics

1 answer:

katen-ka-za [31]2 years ago

3 0

Answer:

RPQ = 239°

Step-by-step explanation:

Since SP is a straight line going through the center of a circle, it is a diameter.

We can say that m<SOR and m<ROP are supplementary and add up to 180° because they form a straight line. We can set up an equation:

m<SOR + m<ROP = 180°

We can substitute in the value of m<SOR:

31° + m<ROP = 180°

m<ROP = 149°

Next, we can also say that m<SOQ and m<QOP are supplementary because they form a straight line. Also, since QO is perpendicular to SP, we can say that both m<SOQ and m<QOP equal to 90°.

Now, we can say that m<ROQ (reflex angle) is equal to the sum of m<QOP and m<ROP from angle addition postulate. We can write the equation:

m<ROQ = m<QOP + m<ROP

m<ROQ = 90° + 149° = 239°

The reflex angle <ROQ cuts the arc RPQ, so they would have the same measure. So, arc RPQ = 239°

You might be interested in

Use spherical coordinates. Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 81, above the xy-plane, an

Natasha_Volkova [10]

Answer:

The volume of the solid is 243 $\sqrt{2} \ \pi$

Step-by-step explanation:

From the information given:

BY applying sphere coordinates:

0 ≤ x² + y² + z² ≤ 81

0 ≤ ρ² ≤ 81

0 ≤ ρ ≤ 9

The intersection that takes place in the sphere and the cone is:

$x^2 +y^2 ( \sqrt{x^2 +y^2 })^2 = 81$

$2(x^2 + y^2) =81$

$x^2 +y^2 = \dfrac{81}{2}$

Thus; the region bounded is: 0 ≤ θ ≤ 2π

This implies that:

$z = \sqrt{x^2+y^2}$

ρcosФ = ρsinФ

tanФ = 1

Ф = π/4

Similarly; in the X-Y plane;

z = 0

ρcosФ = 0

cosФ = 0

Ф = π/2

So here; $\dfrac{\pi}{4} \leq \phi \le \dfrac{\pi}{2}$

Thus, volume: $V = \iiint_E \ d V = \int \limits^{\pi/2}_{\pi/4} \int \limits ^{2\pi}_{0} \int \limits^9_0 \rho ^2 \ sin \phi \ d\rho \ d \theta \ d \phi$

$V = \int \limits^{\pi/2}_{\pi/4} \ sin \phi \ d \phi \int \limits ^{2\pi}_{0} d \theta \int \limits^9_0 \rho ^2 d\rho$

$V = \bigg [-cos \phi \bigg]^{\pi/2}_{\pi/4} \bigg [\theta \bigg]^{2 \pi}_{0} \bigg [\dfrac{\rho^3}{3} \bigg ]^{9}_{0}$

$V = [ -0+ \dfrac{1}{\sqrt{2}}][2 \pi -0] [\dfrac{9^3}{3}- 0 ]$

V = 243 $\sqrt{2} \ \pi$

4 0

3 years ago

Help 3 and 4 please!!!!!!!!!!!!

aleksandrvk [35]

Easy peasy

the bit where it says
D={something}
those are the numbers you should input for x to get y values

3.
first solve for y to make life easier
-3x-5y=20
times -1
3x+5y=-20
minus 3x both sides
5y=-3x-20
divide both sides by 5
y=-3/5x-4
sub value of the domain
x=-10
y=-3/5(-10)-4
y=6-4
y=2
a point is (-10,2)

x=-5
y=-3/5(-5)-4
y=3-4
y=-1
(-5,-1) is another

x=0
y=-3/5(0)-4
y=-4
(0,-4) is another

x=5
y=-3/5(5)-4
y=-3-4
y=-7
(5,-7)

points are (-10,2), (-5,-1), (0,-4), (5,-7)

4.
input
x=-2
y=(-2)^2-3
y=4-3
y=1
(-2,1)

x=-1
y=(-1)^2-3
y=1-3
y=-2
(-1,-2)

x=0
y=(0)^2-3
y=-3
(0,-3)

x=1
y=(1)^2-3
y=1-3
y=-3
(1,-3)

x=2
y=(2)^2-3
y=4-3
y=1
(2,1)

the points are (-2,1), (-1,-2), (0,-3), (1,-2), (-2,1)

see graph below