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3 years ago

If f(x) = 5x^2– 3 and f(x + a) = 5x^2+ 30x + 42, what is the value of a ?

Mathematics

1 answer:

Andrei [34K]3 years ago

8 0

Answer:

The zeros are x=0,3,-2

There is a multiplicity of 1 for all of them.

Step-by-step explanation:

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Which is an asymptofe of the graph the function y=cot(x-2π/3)

Shalnov [3]

Hello.

C) x=4π/3

The variable x in the cotangent argument has a unit coefficient, so the period is π, just as it is in the parent function cot(x).

Can you graph y = cot(x)? By subtracting the constant π/6 from the argument, that graph is translated to the right by π/6. Just as with cot(x), it is decreasing everywhere.

Have a nice day

5 0

3 years ago

Question 8 (1 point)

ra1l [238]

Probly the first one it looks right

6 0

3 years ago

Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24

Nat2105 [25]

Answer:

E. 16 min 12 sec

Step-by-step explanation:

Let x represent total time taken to complete the exercise.

We have been given that Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session.

When the readout indicated 24 min 18 sec, she had completed 10% of her exercise session. This means that 90% time of exercise is equal to 24 minutes and 18 seconds.

18 seconds will be equal to 0.3 minutes.

Let us find total time of exercise as:

$\frac{90}{100}x=24.3$

$0.90x=24.3$

$\frac{0.90x}{0.90}=\frac{24.3}{0.90}$

$x=27$

To find readout when Dara had completed 40% of her exercise session, we need to find 60% of total time.

$\frac{60}{100}(27)=0.60(27)=16.2$

Since our time in in minutes, so we will convert 0.2 minutes to seconds by multiplying by 60.

$0.20(60)=12$

Therefore, the readout will indicate 16 minutes 12 seconds, when Dara had completed 40% of her exercise session.

8 0

3 years ago

When x = -3 on the graph of y = 2|x| + 3 what will be the y-coordinate?

SCORPION-xisa [38]

Substitute -3 into the equation

y = 2|x| + 3
y = 2 (3) + 3
y = 6 + 3
y = 9

Answer D

5 0

3 years ago

Solve the simultaneous equations 4x+3y=1 4x^2+3xy+y^2=2

Fantom [35]

Notice that the second equation,

4x ² + 3xy + y ² = 2

can be written as

x (4x + 3y) + y ² = 2

and the first equation says 4x + 3y = 1, so this reduces to

x + y ² = 2

Solve the first equation for x :

4x + 3y = 1

4x = 1 - 3y

x = (1 - 3y)/4

Substitute this into the reduced second equation to get a quadratic equation in y, which happens to be easily factorized and solved:

(1 - 3y)/4 + y ² = 2

1 - 3y + 4y ² = 8

4y ² - 3y - 7 = 0

(4y - 7) (y + 1) = 0

4y - 7 = 0 or y + 1 = 0

y = 7/4 or y = -1

Solve for x :

x = (1 - 3 (7/4))/4 or x = (1 - 3 (-1))/4

x = -17/16 or x = 1

So the two solutions are (x, y) = (-17/16, 7/4) and (x, y) = (1, -1).

8 0

2 years ago