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2 years ago

Gail ran 2/3 of a mile for 6 straight days. How many miles did she run after the 6th day?

Mathematics

1 answer:

Andrei [34K]2 years ago

5 0

Answer:

Its either 4 or 4.6

Step-by-step explanation:

I used multiplication.

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Please Help!!! I really Need it! Please let me know if I am right for part A... I need help for part B too!

vlabodo [156]

You are correct for part A, good job! :D

As for part B, I <em>believe</em> the answers are function, nonlinear, and discrete.

Please let me know if I was wrong at any time. I hope this helps!!

4 0

2 years ago

Suppose the roots of the equation 2x^2−5x−6=0 are α and β.Find the quadratic equation with roots 1/α and 1/β.

pochemuha

Answer:

6x² + 5x - 2 = 0

Step-by-step explanation:

Given

2x² - 5x - 6 = 0 ← in standard form

with a = 2, b = - 5, c = - 6, then

sum of roots α + β = - $\frac{b}{a}$ = $\frac{5}{2}$

product of roots = $\frac{c}{a}$ = - 3, then

sum of new roots = $\frac{1}{\alpha }$ + $\frac{1}{\beta }$

= $\frac{\beta+\alpha }{\alpha \beta }$

= $\frac{\frac{5}{2} }{-3}$ = - $\frac{5}{6}$

product of new roots = $\frac{1}{\alpha }$ × $\frac{1}{\beta }$

= $\frac{1}{\alpha\beta }$ = - $\frac{1}{3}$

Hence the required equation is

x² + $\frac{5}{6}$ x - $\frac{1}{3}$ = 0 or

6x² + 5x - 2 = 0 ( multiplying through by 6 )

8 0

3 years ago

Three forces act on a hook. Determine the magnitude of the resultant of the force.

Novay_Z [31]

Use Hooke's law... (just kidding)

Break down each force vector into horizontal and vertical components.

$\vec F_1=(1000\,\mathrm N)(\cos30^\circ\,\vec x+\sin30^\circ\,\vec y)\approx(866.025\,\mathrm N)\,\vec x+(500\,\mathrm N)\,\vec y$

$\vec F_2=(1500\,\mathrm N)(\cos160^\circ\,\vec x+\sin160^\circ\,\vec y)\approx(-1409.54\,\mathrm N)\,\vec x+(513.03\,\mathrm N)\,\vec y$

$\vec F_3=(750\,\mathrm N)(\cos195^\circ\,\vec x+\sin195^\circ\,\vec y)\approx(-724.444\,\mathrm N)\,\vec x+(-194.114\,\mathrm N)\,\vec y$

The resultant force is the sum of these vectors,

$\vec F=\displaystyle\sum_{i=1}^3\vec F_i\approx(-1267.96\,\mathrm N)\,\vec x+(818.916\,\mathrm N)\,\vec y$

and has magnitude

$|\vec F|\approx\sqrt{(-1267.96\,\mathrm N)^2+(818.916\,\mathrm N)^2}\approx1509.42\,\mathrm N$

The closest answer is D.

5 0

2 years ago

Enter the trigonometric equation you would use to solve for x in the following right triangle. Do not solve the equation.

USPshnik [31]

Answer:

x = sine 70° / 8

Step-by-step explanation:

6 0

2 years ago

A storage shed is to be built in the shape of a (closed) box with a square base. It is to have a volume of 150 cubic feet. The c

Alenkinab [10]

Answer:

$C(s) = 6s^2 + \dfrac{1500}{s}\\\text{where s is the side of base.}$

Step-by-step explanation:

We are given the following in the question:

A storage shed is to be built in the shape of a (closed) box with a square base.

Volume = 150 cubic feet

Let s be the edge of square base and h be the height.

Volume of cuboid =

$l\times b\times h$

where l is the length, b is the base and h is the height.

Volume of box =

$s^2h = 150\\\\h = \dfrac{150}{s^2}$

Area of base =

$\text{side}\times \text{side} = s^2$

Cost of concrete for the base = $4

Cost of base($) = $4s^2$

Area of roof =

$\text{side}\times \text{side} = s^2$

Cost of material for the roof = $2

Cost of roof ($) = $2s^2$

Area of 4 walls =

$4\times (sh)\\=4sh$

Cost of material for the side = $2.50

Cost of material of side($) =

$2.50\times 4s(\dfrac{150}{s^2})\\\\=\dfrac{1500}{s}$

Total cost

= Cost of base + Cost of 4 sides + Cost of roof

$C(s) = 4s^2 + \dfrac{1500}{s} + 2s^2\\\\C(s) = 6s^2 + \dfrac{1500}{s}$

is the required cost function.

5 0

3 years ago