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3 years ago

What is the distance between the solutions to the quadratic equation x2 – 7x + 10 = 28

Mathematics

1 answer:

Kryger [21]3 years ago

4 0

Answer:

Step-by-step explanation:

Given

x² - 7x + 10 = 28 ( subtract 28 from both sides )

x² - 7x - 18 = 0 ← in standard form

(x - 9)(x + 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 9 = 0 ⇒ x = 9

x + 2 = 0 ⇒ x = - 2

The distance between the solutions is

| 9 - (- 2) | = | 11 = 11 units

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You have 40 minutes to exercise at the gym, and you want to burn 300 calories total using both

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b = 40-a

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3 years ago

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3 years ago

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aleksklad [387]

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3 years ago

A piece of cardboard is 1 meter by 1/21/2 meter. A square is to be cut from each corner and the sides folded up to make an open-

12345 [234]

Answer:

The maximum dimensions of the box:

Length of the box = 0.788676 m

Breadth of the box = 0.288676 m

Step-by-step explanation:

Original piece of cardboard is a square with sides of length s.

Length of the card board = l = 1 m

Breadth of cardboard = b = 1/2 m = 0.5 m

Squares with sides of length x are cut out of each corner of a rectangular cardboard to form a box.

Now, length of the box = L = 1 - 2x

And breadth of the box = B = 0.5 - 2x

Height of the box ,H = x

Volume of the box ,V= L × B × H

$\frac{dV}{dx}=\frac{(1 -2x)(0.5-2x)x}{dx}$

$\frac{dV}{dx}=\frac{d(0.5x-3x^2+4x^3)}{dx}$

$\frac{dV}{dx}=0.5-6x+12x^2$

$\frac{dV}{dx}=O$

x = 0.394338 , 0.105662

$\frac{d^2V}{(dx)^2}=-6+24x$

When , $x=0.105662$ , $\frac{d^2V}{(dx)^2}$ (maxima)

The maximum dimensions of the box:

Length of the box = L = 1 - 2x = 1 - 2(0.105662) m = 0.788676 m

Breadth of the box = B = 0.5 - 2x = 0.5 - 2(0.105662) m = 0.288676 m

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3 years ago

(x+9)^2 = -64 ? How to solve this ?

victus00 [196]

X=-9+8i,-9-8i or no solution (i see someone already answered this but this is here just in case)

8 0

4 years ago

Read 2 more answers