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2 years ago

The table shows the solution to the equation|2x-5|-2=3: which is the first step

1 answer:

5 0

|2x - 5| - 2 = 3 |add 2 to both sides

|2x - 5| = 5 ⇔ 2x - 5 = 5 or 2x - 5 = -5 |add 5 to both sides

2x = 10 or 2x = 0 |divide both sides by 2

x = 5 or x = 0

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On a certain hot Summer's day 388 people use the public swimming pool. The daily prices are $1.75 for children and $2 for adults

Alecsey [184]

Answer:

it could be 372 adults with one kid or 426 kids with no adults but that isn't logical

Step-by-step explanation:

i don't really know how to explain it sorry

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2 years ago

What value of x would make the figure a parallelogram (show your math please)

Verdich [7]

Answer:

Step-by-step explanation:

Diagonals of a parallelogram bisects each other.

Therefore,

4x = 8

x = 8/4

x = 2

5x = 3x + 4

5x - 3x = 4

2x = 4

x = 4/2

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8 0

2 years ago

A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,

algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}$

Where $v_i$ is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is $[0, \pi/2]$ . The critical points of the function are those who make $d'(\theta)=0$ :

$d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}$

$d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...$

The critical value inside the interval is $\pi/4$ .

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft$

The second step is to find the values of the function at the endpoints of the interval:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft$

The biggest value of f is gived by $\pi/4$ , therefore $\pi/4$ is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

4 0

2 years ago

Can someone please help me?

Luda [366]

Answer:

(x-1)^5 (x+1)

Step-by-step explanation:

2 (x-1)^5 + (x-1)^6

Factor out (x-1)^5

2 (x-1)^5 + (x-1)^5 (x-1)

(x-1)^5( 2 + x-1)

Combine like terms

(x-1)^5 (x+1)

4 0

2 years ago

Read 2 more answers

Determine an ordered pair that is a solution to this function ? Y=1.9x+56

In-s [12.5K]

(0, 56)

hope this helps :D

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2 years ago