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Serga [27]
2 years ago
8

Given the function f(x)= 3(x-2)² + 5, what is f(-1)?

Mathematics
2 answers:
motikmotik2 years ago
5 0

Answer:

32

Step-by-step explanation:

f(x)= 3(x-2)² + 5

Let x = -1

f(-1)= 3(-1-2)² + 5

Parentheses first

f(-1) = 3(-3)^2 +5

Exponents

f(-1)= 3*9 + 5

Multiply

f(-1) = 27+5

Add

f(-1) = 32

Anna11 [10]2 years ago
5 0
Answer: 32

=3(-1-2)^2+5
=3(-3)^2+5
=3•3^2+5
=3^3+5
=27+5
=32

Explanation: Plug in -1 for x in the function and solve.
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What is the solution to the equation x-16=-8
Aneli [31]

Answer:

x = 8

Step-by-step explanation:

8 minus 16 equals to a negative answer, resulting to -8.

5 0
2 years ago
Read 2 more answers
Variable y varies directly with x2, and y = 96 when x = 4.
lakkis [162]
Direct variation is y = kx, where k is the constant of variation.
 But now it says y varies directly with x2 (or 2x), so now the x in the equation is 2x.

The equation is y = k(2x)
 Now you find k.
y = 96 when x = 4.
(96) = k(2*4)
96 = k(8)
k = 12
 
The equation is now y = 12(2x)
To find the value of y when x=2, plug 2 into the equation you made. y = 12(2*2) y = 48
_________________
Now it's with a "quadratic variation," which is the same thing except x is squared.
The equation is y = kx^2
 
But y varies directly with x2 (same thing as 2x), so now it's y = k(2x)^2.
 
Now you find k by substituting y and x values that were given.
y = 180 when x = 6
(180) = k(2*6)^2
180 = k(12)^2
180 = k(144)
k = 1.25
k, 1.25, is the constant of variation.
4 0
2 years ago
Consider a propeller driven aircraft in forward flight at a speed of 150 mph: The propeller is set up with an advance ratio such
meriva

a)  The thrust of the propeller is  23,834 lbf

b)  The induced velocity is 8.44 mph

c)  The velocity in the downstream far field away from the propeller disc     is 161.56 mph

d) The power absorbed by the fluid is  42.7 hp

e)  the propeller power (thrust flight speed) is  4,844.08 hp

f)  The propeller efficiency is 113.6%

a)The thrust of the propeller can be calculated using the equation,

Thrust = 2πρR^2V^2, where ρ is the air density, R is the propeller radius, and V is the average velocity of the propeller disc. For sea level at a speed of 150 mph, the air density is about 0.002377 slugs/ft^3.

the thrust of the propeller can be calculated as: Thrust = 2π x 0.002377 x (6 ft)^2 x (170 mph)^2 = 23,834 lbf

b)The induced velocity can be calculated using the equation v_ind = (1/2) V ∞ [1–(V_∞/V_tip)^2]^(1/2), where V_∞ is the free stream velocity and V_tip is the tip velocity of the propeller. For the given problem, the induced velocity can be calculated as: v_ind = (1/2) x 150 mph x [1–(150 mph/170 mph)^2]^(1/2) = 8.44 mph

c) The velocity in the downstream far field can be calculated using the equation V_∞ = V_tip – v_ind, where V_tip is the tip velocity of the propeller and v_ind is the induced velocity. For the given problem, the velocity in the downstream far field can be calculated as: V_∞ = 170 mph – 8.44 mph = 161.56 mph

d)The power absorbed by the fluid can be calculated using the equation P_fluid = (1/2) ρ V_ind^3 A_disc, where ρ is the air density, V_ind is the induced velocity, and A_disc is the area of the propeller disc. For the given problem, the power absorbed by the fluid can be calculated as P_fluid = (1/2) x 0.002377 x (8.44 mph)^3 x (π x (6 ft)^2) = 42.7 hp

e)The power absorbed by the propeller can be calculated using the equation P_prop = T x V, where T is the thrust of the propeller and V is the flight speed of the aircraft. For the given problem, the power absorbed by the propeller can be calculated as: P_prop = 23,834 lbf x 150 mph = 3,575,500 ft-lbf/s = 4,844.08 hp

f)The efficiency of the propeller can be calculated using the equation η = P_prop/P_fluid, where P_prop is the power absorbed by the propeller and P_fluid is the power absorbed by the fluid. For the given problem, the efficiency of the propeller can be calculated as: η = 4,844.08 hp/42.7 hp = 113.6%

To know  more  about induced velocity refer to the link brainly.com/question/15280442

#SPJ4

8 0
1 year ago
there are 10 crates of eggs stacked in the corner. each crate of eggs holds 20 eggs. If there is only 1 broken egg in the entire
alisha [4.7K]

Answer: 5%

Step-by-step explanation: If there is one broken egg in each crate (1/20), you would change that to5%

And if there are ten crates, then you see how many eggs there a re total.

(10 × 20 = 200)

If there are 200 eggs and for every 20 eggs there is on broken one, then there will be 10 broken eggs total. or 10/200

convert the fraction to a decimal ( 10 ÷ 200 = .05)

then convert the decimal to a percent. .05 is equal to 5%

PLEAZE RATE BRAINLIEST!!!

3 0
2 years ago
Solve the equation.
Nutka1998 [239]

Answer:

C

Step-by-step explanation:

In this technique, if we have to factorise an expression like ax2+bx+c, we need to think of 2 numbers such that:

N1⋅N2=a⋅c=1⋅−12=−12

AND

N1+N2=b=−1

After trying out a few numbers we get N1=3 and N2=−4

3⋅−4=−12, and 3+(−4)=−1

x2−x−12=x2−4x+3x−12

x(x−4)+3(x−4)=0

(x+3)(x−4)=0

Now we equate the factors to zero.

x+3=0,x=−3

x−4=0,x=4

3 0
2 years ago
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