PLEASE HELP

What is m∠PTR? a. 12 b. 40 c. 50 d. 140

nata0808 [166]

Answer:

d. 140

Step-by-step explanation:

QTR is a 180 degree angle, so we can assume PTR is 140 degrees because (x+28) is (12+28)= 40

7 0

3 years ago

marshall took $36.75 to a fair. Each ticket into the fair costs X dollars. marshall bought 3 tickets. which expression represent

Sholpan [36]

Y = money left over
X = fair ticket

Y = 36.75 - 3x

7 0

3 years ago

HELP PLS! A boat is heading towards a lighthouse, whose beacon-light is 102 feet above the water. From point AA, the boat’s crew

KatRina [158]

Answer:

Distance from point AA to point BB = 189.3 feet

Step-by-step explanation:

Let the distance from point AA to the base of the lighthouse be represented by x, and the distance from point BB to the base of the lighthouse be represented by y. So that;

distance from point AA to point BB = x - y

To determine the value of x, applying the required trigonometric function;

Tan θ = $\frac{opposite}{sdjacent}$

Tan 13 = $\frac{102}{x}$

x = $\frac{102}{Tan 13}$

= 441.81 feet

x = 441.8 feet

To determine the value of y;

Tan 22 = $\frac{102}{y}$

y = $\frac{102}{Tan 22}$

= 252.46

y = 252.5 feet

Thus,

distance from point AA to point BB = 441.8 - 252.5

= 189.3 feet

5 0

3 years ago

1) Which of the following shows the correct steps to find the value of 27 1/3?

sesenic [268]

First of all, we need to know: 3^3 = 3 x 3 x 3 = 9 x 3 = 27

So:

$(27)^{\frac{1}{3}} = (3^3)^{\frac{1}{3}} = (3)^{3\times \frac{1}{3}} = 3^{1} = 3$

So, your answer is D.

4 0

4 years ago

Read 2 more answers

g 1.32 Two points on a sphere of radius 3 are given as P1(3,0,30) and P2(3,45,45): (a) Find the position vectors of P1 and P2. (

Rama09 [41]

Answer:

a) P.V of is OP₁ = [ 1.5i + 0j + 2.6k ], P.V of is OP₂ = [ 1.5i + 1.5j + 2.12k ]

b) Vector connecting P₁ to P₂ is [ 0i + 1.5j + 0.48k ]

c) cylindrical coordinates are (1.5, π/2, 0.48)

Step-by-step explanation:

Given that;

r = 3

P₁ ( 3, 0°, 30° ), P₂ ( 3, 45°, 45° )

a)

P.V of P₁

x = rcos∅sin∅ = 3(cos0°) ( sin30°) = (3 × 1 × 0.5) = 1.5

y = rsin∅sin∅ = 3(sin0°) (sin30°) = (3 × 0 × 0.5) = 0

z = rcos∅ = 3(cos30°) = ( 3 × 0.866) = 2.6

∴ P.V of is OP₁ = [ 1.5i + 0j + 2.6k ]

P.V of P₂

x = rcos∅sin∅ = 3(cos45°) ( sin45°) = (3 × 0.7071 × 0.7071) = 1.5

y = rsin∅sin∅ = 3(sin45°) (sin45°) = (3 × 0.7071 × 0.7071) = 1.5

z = rcos∅ = 3(cos45°) = ( 3 × 0.7071) = 2.12

∴ P.V of is OP₂ = [ 1.5i + 1.5j + 2.12k ]

b)

Vector connecting P₁ to P₂ is given by

OP₂ - OP₁ = [ 1.5i + 1.5j + 2.12k ] - [ 1.5i + 0j + 2.6k ]

= [ 0i + 1.5j + 0.48k ]

c)

P₁P₂ → = [ 0i + 1.5j + 0.48k ] = [ 1.5j + 0.48k ]

so in a cylindrical coordinate, it should be

r = √(o² + 1.5²) = 1.5

∅ = tan⁻¹[y/π] = π/2

z = 0.48

cylindrical coordinates are (1.5, π/2, 0.48)

5 0

3 years ago