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2 years ago

Blue Sky Jeans Company manufactures jeans. A machine cuts the fabric

Mathematics

1 answer:

Sindrei [870]2 years ago

6 0

Answer:

79687

Step-by-step explanation:

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Distributive Property Assignment 1/6(3/5x+18)= Leave answer as a fraction.

Rashid [163]

Answer:

1/10x+3

Step-by-step explanation:

1/6(3/5x+18)

Multiply 3/5x and 18 by 1/6

1/10x+3

6 0

3 years ago

13....................

Mashcka [7]

13.

(a)

$y=-5x^2+21x-3$

for x = 2:

$y=-5(2)^2+21(2)-3=-20+42-3=19$

for x = 5

$y=-5(5)^2+21(5)-3=-125+105-3=-23$

(b)

The graph is:

(c)

Using the graph:

$\begin{gathered} x=1.8,y=18.6 \\ x=0.548,y=7 \\ x=3.65,y=7 \end{gathered}$ $\begin{gathered} x=1.8 \\ y=-5(1.8)^2+21(1.8)-3=-16.2+37.8-3=18.6 \end{gathered}$ $\begin{gathered} y=7 \\ -5x^2+21x-3=7 \\ -5x^2+21x-10=0 \end{gathered}$

Using the quadratic formula:

$\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-21\pm\sqrt[]{21^2-4(-5)(-10)}}{2(-5)} \\ x=\frac{-21\pm\sqrt[]{241}}{-10} \\ x=0.54758 \\ or \\ x=3.6524 \end{gathered}$

6 0

8 months ago

A man is standing at a radar base and observes an unidentified plane at an altitude 6000m flying towards the radar base at an an

liubo4ka [24]

Answer:

The speed in of the plane is 115.47 m/sec

Step-by-step explanation:

Given:

Height at which the plane is flying = 6000 m

Angle of elevation at the radar base = 30 Degrees

Angle of elevation at the radar base after one minute = 60 Degrees

To Find:

The Speed of the plane in meter per second = ?

Solution:

Let us use the tangent of the angle to find the distance (d) to a point directly below plane:

when the angle is 30 degrees

$tan(30) = \frac{6000}{d1}$

$d1 = \frac{6000}{tan(30)}$

$d1 = \frac{6000}{0.577}$

d1 = 10392.3 meters

when the angle is 60 degrees

$tan(60) = \frac{6000}{d2}$

$d2 = \frac{6000}{tan(60)}$

$d2 = \frac{6000}{1.732}\\$

d2 = 3464.1 meters

distance travelled by aircraft in 1 min is

=>d1 - d2

=>0392.3 - 3464.1

= 6928.2 m/min

Now converting to m/sec

=> $\frac{6928.2}{60}$

=>115.47 m/sec

4 0

2 years ago

You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along

sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

$d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}$

Then, the time (speed divided by distance) is:

$t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1$

To optimize this function we have to derive and equal to zero:

$\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\$

$x$

As $d_b=x$ , the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

7 0

3 years ago

WILL GIVE BRAINLIEST

Alex

Answer:

Step-by-step explanation:

(3.14)(15*15)(315)=222548

5 0

3 years ago