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2 years ago

What are the combining like terms plz

Mathematics

2 answers:

Mekhanik [1.2K]2 years ago

8 0

Answer:

-7q and 9q, 2 and 6

Step-by-step explanation:

they are like each other

Lynna [10]2 years ago

6 0

Answer: Terms that have identical variable parts (same variable(s) and same exponent(s)). So in this case the like terms are -7q and 9q AND 2 and 6 are like terms *REMEMBER: constant (which are numbers with NO variables can also be like terms)

but to solve this:

first group the like terms so:

(-7q+9q) that gives us 2q

(2+6) that gives us 8

*In your FINAL answer ALWAYS put the number(s) with variables first so the final answer is 2q+8

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The top of the Boulder Dam has an angle of elevation of 1.2 radians from a point on the Colorado River. Measuring the angle of e

Nana76 [90]

Answer:

382.925 feets

Step-by-step explanation:

The solution diagram is attached below :

Converting radian measurement to degree :

radian angle * 180/π = degree angle

1.2 * 180/π = 68.755°

0.9 * 180/π = 51.566°

Height of dam is h:

Using trigonometry :

Tan θ = opposite / Adjacent

Tan 68.755° = h / x

h = x Tan 68.755° - - - (1)

Tan 51.566° = h / (155+x)

h = (155+x) tan 51.566° - - - (2)

Equate (1) and (2)

x Tan 68.755 = (155+x) Tan 51.566

x Tan 68.755 = 155tan 51.566 + x tan 51.566

x Tan 68.755 = 195.32311 + x Tan 51.566

x Tan 68.755 - x Tan 51.566 = 195.32311

x(tan 68.755 - tan 51.566) = 195.32311

x * 1.3120110 = 195.32311

1.3120110x = 195.32311

x = 195.32311 / 1.3120110

x = 148.87307

Using :

h = x Tan 68.755

h = 148.87307 * tan(68.755)

h = 382.92539

h = 382.925 feets

5 0

2 years ago

Kesha threw her baton up in the air from the marching band platform during practice. The equation h(t) = −16t² + 54t + 40 gives

lapo4ka [179]

Answer:

a) 40 feet

b) 54 ft/min

c) 4 mins

Step-by-step explanation:

Solution:-

- Kesha models the height ( h ) of the baton from the ground level but thrown from a platform of height hi.

- The function h ( t ) is modeled to follow a quadratic - parabolic path mathematically expressed as:

h ( t ) = −16t² + 54t + 40

Which gives the height of the baton from ground at time t mins.

- The initial point is of the height of the platform which is at a height of ( hi ) from the ground level.

- So the initial condition is expressed by time = 0 mins, the height of the baton h ( t ) would be:

h ( 0 ) = hi = -16*(0)^2 + 54*0 + 40

h ( 0 ) = hi = 0 + 0 + 40 = 40 feet

Answer: The height of the platform hi is 40 feet.

- The speed ( v ) during the parabolic path of the baton also varies with time t.

- The function of speed ( v ) with respect to time ( t ) can be determined by taking the derivative of displacement of baton from ground with respect to time t mins.

v ( t ) = dh / dt

v ( t )= d ( −16t² + 54t + 40 ) / dt

v ( t )= -2*(16)*t + 54

v ( t )= -32t + 54

- The velocity with which Kesha threw the baton is represented by tim t = 0 mins.

Hence,

v ( 0 ) = vi = -32*( 0 ) + 54

v ( 0 ) = vi = 54 ft / min

Answer: Kesha threw te baton with an initial speed of vo = 54 ft/min

- The baton reaches is maximum height h_max and comes down when all the kinetic energy is converted to potential energy. The baton starts to come down and cross the platform height hi = 40 feet and hits the ground.

- The height of the ball at ground is zero. Hence,

h ( t ) = 0

0 = −16t² + 54t + 40

0 = -8t^2 + 27t + 20

- Use the quadratic formula to solve the quadratic equation:

$t = \frac{27+/-\sqrt{27^2 - 4*8*(-20)} }{2*8}\\\\t = \frac{27+/-\sqrt{1369} }{16}\\\\t = \frac{27+/-37 }{16}\\\\t = \frac{27 + 37}{16} \\\\t = 4$