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2 years ago

What is the following quotient? sqrt 6 + sqrt 11 / sqrt 5 +sqrt 3

Mathematics

2 answers:

Nadya [2.5K]2 years ago

6 0

The answer is B) Link below

just olya [345]2 years ago

5 0

Answer:

$\frac{\sqrt6 +\sqrt{11}}{\sqrt{5} + \sqrt{3}} \times \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \frac{\sqrt{30} - \sqrt{18} + \sqrt{55} - \sqrt{33}}{2}$

Step-by-step explanation:

Let's translate what you've written in words to an expression with mathematical symbols.

$\frac{\sqrt6 +\sqrt{11}}{\sqrt{5} + \sqrt{3}}$

To solve this, you need to multiply both the numerator and denominator with the conjugate of the denominator.

$\frac{\sqrt6 +\sqrt{11}}{\sqrt{5} + \sqrt{3}} \times \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \frac{\sqrt{30} - \sqrt{18} + \sqrt{55} - \sqrt{33}}{2}$

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Antonio's toy boat is bobbing in the water under a dock. The vertical distance HHH (in \text{cm}cmstart text, c, m, end text) be

musickatia [10]

Answer: Time t = 33.0 seconds

Step-by-step explanation:

Given that the vertical distance H between the dock and the top of the boat's mast t seconds after its first peak is modeled by the function

H(t) = 5cos( 2π/3 t) − 35.5H

Where the maximum vertical distance = 5

At the down position, H(t) = 0

5cos( 2π/3 t) − (35.5/100)H = 0

5cos( 2π/3 t) − 0.355 × 5 =0

5cos( 2π/3 t) − 0.1775 = 0

5cos( 2π/3 t) = 0.1775

cos( 2π/3 t) = 0.1775/5

cos( 2π/3 t) = 0.355

2π/3 t = cos^-1 (0.355)

2π/3 t = 69.2

2πt = 69.2 × 3

2πt = 207.6

t = 207.6/2π

t = 33.0 seconds

3 0

2 years ago

Read 2 more answers

Determine a series of transformations that would map polygon ABCDE onto polygon A'B'C'D'E'?

vovangra [49]

Answer:

A reflection followed by a translation

Step-by-step explanation:

A reflection across they x-axis would put all the points in the right section and on the right y value. The x value would need to be changed though. So, you would need to do a translation of 4 on all points.

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2 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$