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3 years ago

Find the number of solutions of -x^2 + 5x - 4 = 0.

Mathematics

1 answer:

Mekhanik [1.2K]3 years ago

6 0

Answer:

There are 2 solutions for the equation -x^2 + 5x - 4 = 0 i.e x=1 and x=4

Step-by-step explanation:

We can find the number of solutions of the given equation -x^2 + 5x - 4 = 0 by solving the equation using factors method to solve the quadratic equation.

$-x^2 + 5x - 4 = 0\\Taking\,\, - sign\,\, common\\x^2 - 5x + 4 = 0\\making\,\, factors\,\, of\,\, 4x^2\\x^2 -4x -x +4 = 0\\x(x -4) -1 (x - 4) = 0\\(x-1)(x-4)=0\\x-1 = 0 \,\, and \,\, x-4 =0\\x = 1 \,\,and\,\, x = 4\\$

The values of x are:

x=1 and x=4

So, there are 2 solutions for the given equation -x^2 + 5x - 4 = 0

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A student stands 20 m away from the foot

Ostrovityanka [42]

Answer:

Height of tree is $\approx$ <em>15 m.</em>

Step-by-step explanation:

Given that student is 20 m away from the foot of tree.

and table is 1.5 m above the ground.

The angle of elevation is: 34°28'

Please refer to the attached image. The given situation can be mapped to a right angled triangle as shown in the image.

AB = CP = 20 m

CA = PB = 1.5 m

$\angle C$ = 34°28' = 34.46°

To find TB = ?

we can use trigonometric function tangent to find TP in right angled $\triangle TPC$

$tan \theta = \dfrac{Perpendicular}{Base}\\tan C= \dfrac{PT}{PC}\\\Rightarrow tan 34.46^\circ = \dfrac{PT}{20}\\\Rightarrow PT = 20 \times 0.686 \\\Rightarrow PT = 13.72\ m$

Now, adding PB to TP will give us the height of tree, TB

Now, height of tree TB = TP + PB

TB = 13.72 + 1.5 = 15.22 $\approx$ <em>15 m</em>

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4 years ago

What is the domain of the function.....<br> {TIMED PLS HURRY}

timofeeve [1]

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3 years ago

Charlie wants to order lunch for his friends. He’ll order 6 sandwich’s and a $3 kids meal for his little brother Charlie had $27

Vera_Pavlovna [14]

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3 years ago

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The oblique pyramid has a square base. What is the volume of the pyramid? 2.5 cm3 5 cm3 6 cm3 7.5 cm3

Olenka [21]

Take a look at the attachment below. It fills in for the attachment that wasn't provided in the question -

An oblique pyramid is one that has a top not aligned with the base. Due to this, the height of the pyramid connects with two vertices at its ends to form a right angle present outside the pyramid, knowing that it is always perpendicular to the base. There is no difference between the calculations of the volume of an oblique pyramid and a pyramid however -

$\\Base Area = 2 cm * 2 cm = 4 cm^2,\\Volume ( Pyramid ) = 1 / 3 * ( Base Area ) * ( Height ),\\Volume = 1 / 3 * ( 4 ) * ( 3.75 ),\\-------------------------\\Volume = 5 cm^3$

<u><em>And thus, you're solution is 5 cm^3, or in other words option b!</em></u>

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3 years ago

Pre-Calc: Find all the zeros of the function.

uysha [10]

The zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i and the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

<h3>What are polynomial expressions?</h3>

Polynomial expressions are mathematical statements that are represented by variables, coefficients and operators

<h3>How to determine the zeros of the polynomial?</h3>

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Set the equation to 0

So, we have

(5y - 4)(5y + 4)(25y^2 + 16) = 0

Expand the equation

So, we have

5y - 4 = 0, 5y + 4 = 0 and 25y^2 + 16 = 0

This gives

5y = 4, 5y = -4 and 25y^2 = -16

Solve the factors of the equation

So, we have

y = 4/5, y = -4/5 and y = ±4/5√i

Hence, the zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i

How to write the polynomial as a product of the linear factors?

In (a), we have

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Hence, the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)