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3 years ago

What is the solution to the equation 5/3b^3-2b^2-5 = 2/b^3-2?

Mathematics

1 answer:

sweet [91]3 years ago

7 0

Answer:

The solution to the equation is:

Option c. b=0 and b=4

Step-by-step explanation:

5 / (3b^3-2b^2-5) = 2 / (b^3-2)

Cross multiplication:

5(b^3-2)=2(3b^3-2b^2-5)

Applying distributive property both sides of the equation to eliminate the parentheses:

5(b^3)-5(2)=2(3b^3)-2(2b^2)-2(5)

Multiplying:

5b^3-10=6b^3-4b^2-10

Passing all the terms to the right side of the equation: Subtracting 5b^3 and adding 6 both sides of the equation:

5b^3-10-5b^3+10=6b^3-4b^2-10-5b^3+10

Adding like terms:

0=b^3-4b^2

b^3-4b^2=0

Getting common factor b^2 on the left side of the equation:

b^2 (b^3/b^2-4b^2/b^2)=0

b^2 (b-4) = 0

Two solutions:

(1) b^2=0

Solving for b: Square root both sides of the equation:

sqrt(b^2)=sqrt(0)

Square root:

b=0

(2) b-4=0

Solving for b: Adding 4 both sides of the equation:

b-4+4=0+4

Adding like terms:

b=4

The solution of the equation is: b=0 and b=4 (Option c)

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X by itself is 1, otherwise it is the number to the left

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Find the area of triangle ABC. Give your answer correct to 1 decimal place AC=7 Cb=12 a=72 degrees b=59 degrees

Tems11 [23]

Answer:

31.7 cm²

Step-by-step explanation:

The area (A) of the triangle is calculated as

A = $\frac{1}{2}$ absinC

where a, b are the 2 sides given and C the angle between them

∠ C = 180° - (72 + 59)° = 180° - 131° = 49° , thus

A = 0.5 × 12 × 7 × sin49° ≈ 31.7 cm² ( to 1 dec. place )

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3 years ago

A riverboat travels 54 km downstream in 2 hours. It travels 57 km upstream in 3 hours. Find the speed of the boat and the speed

irakobra [83]

Here we must write and solve a system of equations to find the speed of the boat and the speed of the stream. We will find that the boat's speed is 23km/h and the river's speed is 4km/h.

First, remember the relation:

Distance = Speed*Time.

Now let's define the variables we will be using:

B = boat's speed
R = river's speed.

When the boat travels downstream, the total speed of the boat is the speed of the boat plus the speed of the river, and we know that in that case it travels 54km in 2 hours, then:

54km = (B + R)*2h

When the boat travels upstream, we must subtract the speed of the river. In that case, we know that the boat travels 57km in 3 hours, then we have:

57km = (B - R)*3h

Then our system of equations is:

54km = (B + R)*2h

57km = (B - R)*3h

To solve this, first, we need to isolate one of the variables in one of the equations, let's isolate B in the second one:

57km = (B - R)*3h

57km/3h + R = B

19 km/h + R = B

Now we can replace that in the other equation:

54km = (B + R)*2h

54km/2h = B + R

27 km/h = B + R = (19 km/h + R) + R

Now we can solve this for R:

27 km/h = 19km/h + 2*R

27 km/h - 19km/h = 2*R

8 km/h = 2*R

(8km/h)/2 = 4km/h = R

Now that we know the value of R, we can use:

B = 19 km/h + R = 19 km/h + 4km/h = 23 km/h

So the boat's speed is 23km/h and the river's speed is 4km/h.

If you want to learn more, you can read:

brainly.com/question/12895249

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3 years ago

Solve the following quadratic equation using the quadratic formula. Which of the following expressions gives the numerators of t

Illusion [34]

I hope the choices for the numerators of the solutions are given.

I am showing the complete work to find the solutions of this equation , it will help you to find an answer of your question based on this solution.

The standard form of a quadratic equation is :

ax² + bx + c = 0

And the quadratic formula is:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

So, first step is to compare the given equation with the above equation to get the value of a, b and c.

So, a = 10, b = -19 and c = 6.

Next step is to plug in these values in the above formula. Therefore,

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So, $x=\frac{19-11}{20} ,\frac{19+11}{20}$

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What term is 1/1024 in the geometric sequence,-1,1/4,-1/6..?

Trava [24]

Answer:

$\large\boxed{\text{sixth term is equal to}\ \dfrac{1}{1024}}$

Step-by-step explanation:

The explicit formula for a geometric sequence:

$a_n=a_1r^{n-1}$

$a_n$ - n-th term

$a_1$ - first term

$r$ - common ratio

$r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=...=\dfrac{a_n}{a_{n-1}}$

We have

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If $a_3=-\dfrac{1}{16}$ then the common ratio is $r=\dfrac{-\frac{1}{16}}{\frac{1}{4}}=-\dfrac{1}{16}\cdot\dfrac{4}{1}=-\dfrac{1}{4}$

Put to the explicit formula:

$a_n=-1\left(-\dfrac{1}{4}\right)^{n-1}$

Put $a_n=\dfrac{1}{1024}$ and solve for <em>n </em>:

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$\dfrac{1}{4^n}=\dfrac{1}{4^6}\iff n=6$

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3 years ago