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evablogger [386]
3 years ago
8

Suppose a and b are the solutions to the quadratic equation 2x^2-3x-6=0. Find the value of (a+2)(b+2).

Mathematics
1 answer:
KonstantinChe [14]3 years ago
3 0

Answer:

(a+2)(b+2) = 4

Step-by-step explanation:

We are given the following quadratic equation:

2x^2-3x-6=0

Let a a and b be the solution of the given quadratic equation.

Solving the equation:

2x^2-3x-6=0\\\text{Using the quadratic formula}\\\\x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\text{Comparing the equation to }ax^2 + bx + c = 0\\\text{We have}\\a = 2\\b = -3\\c = -6\\x = \dfrac{3\pm \sqrt{9-4(2)(-6)}}{4} = \dfrac{3\pm \sqrt{57}}{4}\\\\a = \dfrac{3+\sqrt{57}}{4}, b = \dfrac{3-\sqrt{57}}{4}

We have to find the value of (a+2)(b+2).

Putting the values:

(a+2)(b+2)\\\\=\bigg(\dfrac{3+\sqrt{57}}{4}+2\bigg)\bigg(\dfrac{3-\sqrt{57}}{4}+2\bigg)\\\\=\bigg(\dfrac{11+\sqrt{57}}{4}\bigg)\bigg(\dfrac{11-\sqrt{57}}{4}\bigg)\\\\=\dfrac{121-57}{156} = \dfrac{64}{16} = 4

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In ∆MNP, m∠N = 90º, NH – altitude, m∠P = 21º, PM = 4 cm. Find MH.
Sveta_85 [38]

Answer:

0.51 cm

Step-by-step explanation:

In right triangle MNP, MP = 4 cm, m∠N = 90°, m∠P = 21°

By the sine definition,

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Now, consider right triangle HMN (it is right because NH is an altitude). By the cosine definition,

\cos \angle M=\dfrac{\text{Adjacent leg}}{\text{Hypotenuse}}=\dfrac{MH}{MN}\\ \\MH=MN\cos \angle M

In the right triangle, two acute angles are always complementary, so

m\angle M=90^{\circ}-m\angle P=90^{\circ}-21^{\circ}=69^{\circ}

Thus,

MH=1.43\cos 69^{\circ}\approx 0.51\ cm

7 0
3 years ago
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Blizzard [7]

Answer:

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Step-by-step explanation:

first divide -2 by 5= -2/5

multiply 18 by -2/5=-36/5

divide 36 by 5=7.2

4 0
3 years ago
Which expression is equivalent to (a^8)^4
crimeas [40]
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5 0
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Solve x'=5t(sqrt(x)) x(0)=1
LekaFEV [45]

Answer:

2\sqrt{x}=\frac{5t^2}{2}+2

Step-by-step explanation:

Given: \frac{\mathrm{d} x}{\mathrm{d} t}=5t\sqrt{x}\,,\, x(0)=1

Solution:

A differential equation is said to be separable if it can be written separately as  functions of two variables.

Given equation is separable.

We can write this equation as follows:

\frac{dx}{\sqrt{x}}=5t\,dt

On integrating both sides, we get

\int \frac{dx}{\sqrt{x}}=\int 5t\,dt

Formulae Used:

\int \frac{1}{\sqrt{x}}=2\sqrt{x}\,\,,\,\,\int t\,dt=\frac{t^2}{2}

So, we get solution as 2\sqrt{x}=\frac{5t^2}{2}+C

Applying condition: x(0) = 1, we get C=2

Therefore, 2\sqrt{x}=\frac{5t^2}{2}+2

8 0
3 years ago
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Answer:

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Step-by-step explanation:

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