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3 years ago

Find the solution of each equation if the replacement sets are a= { 4,5,6,7,8 } and b= { 9,10,11, 12,13 }.

Mathematics

1 answer:

irga5000 [103]3 years ago

4 0

The answer is 8 so the answer is in set a

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5. If I multiply a number by 3 and then subtract 4, the result is the same as twice the number. Find this number.

Lina20 [59]

Answer:

4 is the number.

Step-by-step explanation:

Take the missing number as x,

→ (x × 3) - 4 = (x × 2)

→ 3x - 4 = 2x

→ 3x = 2x + 4

→ 3x - 2x = 4

→ [ x = 4 ]

Hence, the number is 4.

5 0

3 years ago

What are the solutions of the equation x6 + 6x3 + 5 = 0? Use factoring to solve.

lapo4ka [179]

Consider the equation $x^6+6x^3+5=0$ .

First, you can use the substitution $t=x^3$ , then $x^6=(x^3)^2=t^2$ and equation becomes $t^2+6t+5=0$ . This equation is quadratic, so

$D=6^2-4\cdot 5\cdot 1=36-20=16=4^2,\ \sqrt{D}=4,\\ \\ t_{1,2}=\dfrac{-6\pm 4}{2} =-5,-1$ .

Then you can factor this equation:

$(t+5)(t+1)=0$ .

Use the made substitution again:

$(x^3+5)(x^3+1)=0$ .

You have in each brackets the expression like $a^3+b^3$ that is equal to $(a+b)(a^2-ab+b^2)$ . Thus,

$x^3+5=(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25}) ,\\x^3+1=(x+1)(x^2-x+1)$

and the equation is

$(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25})(x+1)(x^2-x+1)=0$ .

Here $x_1=-\sqrt[3]{5} , x_2=-1$ and you can sheck whether quadratic trinomials have real roots:

1. $D_1=(-\sqrt[3]{5}) ^2-4\cdot \sqrt[3]{25}=\sqrt[3]{25} -4\sqrt[3]{25} =-3\sqrt[3]{25}$ .

2. $D_2=(-1)^2-4\cdot 1=1-4=-3$ .

This means that quadratic trinomials don't have real roots.

Answer: $x_1=-\sqrt[3]{5} , x_2=-1$

If you need complex roots, then

$x_{3,4}=\dfrac{\sqrt[3]{5}\pm i\sqrt{3\sqrt[3]{25}}}{2} ,\\ \\x_{5,6}=\dfrac{1\pm i\sqrt{3}}{2}$ .

6 0

4 years ago

1. A student lifts a box of books that weighs 200 N. The box is lifted 2.5 m. How much work

Angelina_Jolie [31]

The student does a work of 500 Newton meter on the box.

Step-by-step explanation:

Given,

Weight of box of books = force = 200 N

Distance the box is lifted = 2.5 m

Work done = Force * distance

Work done = $200*2.5$

Work done = 500 Newton-meter

Newton meter is same as Joule.

Work done = 500 Joules

The student does a work of 500 Newton meter on the box.

Keywords: multiplication, work done

Learn more about multiplication at:

#LearnwithBrainly

7 0

3 years ago

How many cubic meters in 179.66 cubic centimeters

tamaranim1 [39]

Answer:

0.00018

Step-by-step explanation:

formula:Divide the volume by 1e+6

8 0

2 years ago

How to move from fractions as a part to whole to ratios?

elena55 [62]

What do you mean
<span>How to move from fractions as a part to whole to ratios?</span>

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3 years ago