What is 10√273s · 3√14st³u³

If x=6 is the only x-intercept of the graph of a quadratic equation, which statement best describes the discriminant of the equa

Lena [83]

Since x is the x-intercepts, that means x'=x" =6 or in other word the discriminante ( Δ = b² 4ac) =0. Answer is A

3 0

3 years ago

Vanessa draws one side of equilateral AABC on the coordinate plane at points A(-2, 1) and B(4,1). What are the possible coordina

natulia [17]

Answer:

Step-by-step explanation:

Answer E is the only one that is close

7 0

2 years ago

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Help me: 8x13+7x13+6x13=

MAXImum [283]

8•13=104
7•13=91
6•13=78

104+91+78=273

8 0

3 years ago

Read 2 more answers

Set up an equation and solve the following problem.

nordsb [41]

Answer:

The speed of Dave is 42 miles per hour

The speed of Kent is 46 miles per hour .

Step-by-step explanation:

Given as :

The distance cover by Dave = d = 210 miles

The time taken by Dave = t hour

The speed of Dave = s miph

Again

The distance cover by Kent = D = 230 miles

The time taken by Kent = T hour

The speed of Kent = S = (s + 4 ) miph

For Dave

Time = $\dfrac{\textrm Distance}{\textrm Speed}$

So, t = $\dfrac{\textrm d miles}{\textrm s miph}$

Or, t = $\dfrac{\textrm 210 miles}{\textrm s miph}$

For Kent

Time = $\dfrac{\textrm Distance}{\textrm Speed}$

So, T = $\dfrac{\textrm D miles}{\textrm S miph}$

Or, T = $\dfrac{\textrm 230 miles}{\textrm (s + 4) miph}$

∵ Time taken by both is same

So, t = T

Or, $\dfrac{\textrm 210 miles}{\textrm s miph}$ = $\dfrac{\textrm 230 miles}{\textrm (s + 4) miph}$

Or, 210 × (s + 4) = 230 × s

Or, 210 × s + 210 × 4 = 230 × s

Or, 210 × 4 = 230 × s -210 × s

Or, 210 × 4 = 20 × s

∴ s = $\dfrac{840}{20}$

i.e s = 42 miph

So, The speed of Dave = s = 42 miles per hour

Again

The speed of Kent = S = (s + 4 ) miph

i.e S = 42 + 4

or, S = 46 miph

So, The speed of Kent = S = 46 miles per hour

Hence,The speed of Dave is 42 miles per hour

And The speed of Kent is 46 miles per hour . Answer

8 0

2 years ago

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the bin is constructed from 48 square feet of sheet metal, so:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, the dimensions of the largest volume of such a bin is:

Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$

8 0

3 years ago