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3 years ago

How do I solve the problem

Mathematics

1 answer:

Illusion [34]3 years ago

3 0

I may be wrong but I think a senior citizen ticket costs $8.00 and a student ticket costs $14.00

Hope I helped! :)

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Can some one please help me find the solution of the question

creativ13 [48]

Answer:

x=16°

Step-by-step explanation:

tanθ=opposite/adjacent

tan(<CAB)=BC/AB

<CAB=tan⁻¹(21/29)=35.9097308°

tanθ=opposite/adjacent

tan(<DAB)=BD/AB

<DAB=tan⁻¹((21/2)/29)=19.90374954°

<CAB=<DAB+x

x=<CAB-<DAB

=35.9097308-19.90374954

x=16.00598126°

8 0

2 years ago

How many cubic yards of concrete are in a light pole caisson having a diameter of 24 inches and a

lesya [120]

Answer:

C. 0.7 yd³

Step-by-step explanation:

Find the volume of cylinder:

V = πr²h

Convert the dimensions into yards:

d = 24 inches = 24/36 yards = 2/3 yards ⇒ r = 1/2d = 1/2*2/3 = 1/3 yards
h = 6 feet = 6/3 yards = 2 yards

Find the volume by substituting the values:

V = 3.14*(1/3)²*2 = 0.69777 ≈ 0.7 yd³

Correct choice is C

3 0

2 years ago

the expression 12(x+4) represents the total number of CDs Mie bought in April and May at $12 each. Which property is applied to

Georgia [21]

Distributive Property because 12(x+4)= 12x (the first number or letter in the parenthesis times 12) plus 12x4 (the second number in the parenthesis)
So together it is:
12x+(12x4)

4 0

3 years ago

Kevin can travel 22 1/2 miles in 1/3 of an hour what is his average speed in miles per hour

Natasha_Volkova [10]

$\frac{22.5}{ \frac{1}{3} \times 60}$

4 0

3 years ago

Determine the equation of the line that is perpendicular to the lines r(t) equalsleft angle6t,1 plus3 t,4tright angle and R(

Lynna [10]

$\vec r(t)=\langle6t,1+3t,4t\rangle$

$\vec R(s)=\langle2+s,-8+3s,-12+4s\rangle$

Take the derivatives of each to get the tangent vectors:

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=\langle6,3,4\rangle$

$\dfrac{\mathrm d\vec R(s)}{\mathrm ds}=\langle1,3,4\rangle$

Take the cross product of the tangent vectors to get a vector that is normal to both lines:

$\langle6,3,4\rangle\times\langle1,3,4\rangle=\langle0,-20,15\rangle$

The two given lines intersect when $\vec r(t)=\vec R(s)$ :

$\langle6t,1+3t,4t\rangle=\langle2+s,-8+3s,-12+4s\rangle\implies t=1,s=4$

that is, at the point (6, 4, 4).

The line perpendicular to both of the given lines through the origin is obtained by scaling the normal vector found earlier by $\tau\in\Bbb R$ ; translate this line by adding the vector $\langle6,4,4\rangle$ to get the line we want,

$\vec\rho(\tau)=\langle6,4,4\rangle+\langle0,-20,15\rangle\tau$

$\boxed{\vec\rho(\tau)=\langle6,4-20\tau,4+15\tau\rangle}$

6 0

2 years ago