All categories

3 years ago

Simplify fully a) √12b) √27c) √50

Mathematics

2 answers:

xenn [34]3 years ago

3 0

A) 2√3
B)3√3
C)5√2
Because radicals multiply like rational numbers

DIA [1.3K]3 years ago

3 0

A) 2sqrt3
b) 3sqrt3
c) 5sqrt2

You might be interested in

Write the equation of a line that goes through the points (1,-1) and (3,2).

babunello [35]

Answer:

y=3/2x-5/2

Step-by-step explanation:

1.find the slope

y2-y2/x2-x1

2-(-1)/3-1

2+1/3-1

3/2 slope (m)

2.y-y1=m(x-x1)

y-(-1)=3/2(x-1)

y+1=3/2x-3/2

y=3/2x-3/2-1

y=3/2x-5/2

6 0

3 years ago

The rod is made of A-36 steel and has a diameter of 0.22 in . If the rod is 4 ft long when the springs are compressed 0.7 in . a

ziro4ka [17]

The force in the rod when the temperature is 150 °F is 718.72 pounds-force.

<h3>How to determine the resulting the resulting force due to mechanical and thermal deformation</h3>

Let suppose that rod experiments a <em>quasi-static</em> deformation and that both springs have a <em>linear</em> behavior, that is, force ( $F$ ), in pounds-force, is directly proportional to deformation. Then, the elongation of the rod due to <em>temperature</em> increase creates a <em>spring</em> deformation additional to that associated with <em>mechanical</em> contact.

Given simmetry considerations, we derive an expression for the <em>spring</em> force ( $F$ ), in pounds-force, as a sum of mechanical and thermal effects by principle of superposition:

$F = k\cdot (\Delta x + 0.5\cdot \Delta l)$ (1)

Where:

$k$ - Spring constant, in pounds-force per inch.
$\Delta x$ - Spring deformation, in inches.
$\Delta l$ - Rod elongation, in inches.

The <em>rod</em> elongation is described by the following <em>thermal</em> dilatation formula:

$\Delta l = \alpha \cdot L_{o}\cdot (T_{f}-T_{o})$ (2)

Where:

$\alpha$ - Coefficient of linear expansion, in $\frac{1}{^{\circ}F}$ .
$L_{o}$ - Initial length of the rod, in inches.
$T_{o}$ - Initial temperature, in degrees Fahrenheit.
$T_{f}$ - Final temperature, in degrees Fahrenheit.

If we know $k = 1000\,\frac{lb}{in}$ , $\Delta x = 0.7\,in$ , $\alpha = 6.5\times 10^{-6}\,\frac{1}{^{\circ}F}$ , $L_{o} = 48\,in$ , $T_{o} = 30\,^{\circ}F$ and $T_{f} = 150\,^{\circ}F$ , then the force in the rod at final temperature is:

$F = \left(1000\,\frac{lb}{in} \right)\cdot \left[0.7\,in + 0.5\cdot\left(6.5\times 10^{-6}\,\frac{1}{^{\circ}F} \right)\cdot (48\,in)\cdot (150\,^{\circ}F-30\,^{\circ}F)\right]$

$F = 718.72\,lbf$

The force in the rod when the temperature is 150 °F is 718.72 pounds-force. $\blacksquare$

To learn more on deformations, we kindly invite to check this verified question: brainly.com/question/13774755

3 0

3 years ago

17.

Marta_Voda [28]

Answer:

a) 50 meters

b) 2354 meters

c) 12 seconds

d) 24.13 seconds

Step-by-step explanation:

Hello!

<h3>a) How far is Lincoln above the ground when he shoots the gun?</h3>

We need to really understand what the variables represent in this question. Given that t is time, and h is height, the time at which he shoots the bullet will be 0, as the bullet hasn't started traveling yet. This means that we can solve for the original height of the bullet by plugging in 0 for time in the equation.

Equation: $h = -16t^2 + 384t + 50$

Plug in 0 for t:

$h = -16t^2 + 384t + 50$
$h = -16(0)^2 + 384(0) + 50$
$h = 50$

The height at which Lincoln shot the bullet is 50 meters.

We can also look at this graphically. The height of origin will simply be when the x-value (in this case t-value) is 0. That means it is the point at which the graph intersects the y-axis, known as the y-intercept.

Standard form of a Parabola: $y = ax^2 + bx + c$

The y-intercept is the "c" value.

Given our equation: $h = -16t^2 + 384t + 50$

The c-value is 50. This proves that the y-intercept is 50.

<h3 /><h3>b)How high does the bullet travel vertically relative to the ground?</h3>

We want to find the highest point of the graph. To do that, we can find the vertex.

We can utilize the Axis of Symmetry (AOS) to find the Vertex.

First, find the AOS using the formula: $AOS = -\frac{b}{2a}$

a = -16
b = 384

Plug it into the formula:

$AOS = -\frac{b}{2a}$
$AOS = -\frac{384}{2(-16)}$
$AOS = \frac{384}{32}$
$AOS = 12$

Plug in 12 for t in the equation:

$h = -16t^2 + 384t + 50$
$h = -16(12)^2 + 384(12) + 50$
$h = -2304 + 4608 + 50$
$h = 2304+50$
$h = 2354$

Therefore, the highest point of the bullet is 2354 meters.

<h3>c)How long does it take the bullet to reach its greatest height?</h3>

We answered this in the last question. The t-value when h is at its highest is 12.

<h3>d)After how many seconds (round to nearest 100th) does the bullet hit the ground?</h3>

We have to solve for the values of t when h is 0 (when it touches the ground).

Set the equation to 0, and solve using the quadratic formula.

Quadratic formula: $x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$
$x = \frac{-384\pm\sqrt{384^2 - 4(-16)(50)}}{2(-16)}$
$x = \frac{-384\pm\sqrt{147456 +3200}}{-32}$
$x = \frac{-384\pm\sqrt{150656}}{-32}$
$x = \frac{-384\pm388.14}{-32}$
$x = -\frac{4.14}{32}, x = \frac{772.14}{32}$

We are only going to take the positive solution, as we can't have a negative time. The solution that doesn't work is called an extraneous solution.