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Nadusha1986 [10]
3 years ago
7

7X^2/3--55X^1/3--8=0

Mathematics
1 answer:
Neporo4naja [7]3 years ago
5 0

Answer:

x=\frac{2969\cdot \:2801^{\frac{1}{2}}}{686}-\frac{157135}{686},\:x=-\frac{157135}{686}-\frac{2969\cdot \:2801^{\frac{1}{2}}}{686}

Step-by-step explanation:

7x^{\frac{2}{3}}+55x^{\frac{1}{3}}+8=0

7\left(x^{\frac{1}{3}}\right)^2+55x^{\frac{1}{3}}+8=0

Rewrite as if x^{\frac{1}{3}}=u:

7u^2+55u+8=0

Use quadratic equation:

\frac{-55\pm \left(55^2-4\cdot \:7\cdot \:8\right)^{\frac{1}{2}}}{2\cdot \:7}

Solutions for this part:

u=\frac{-55+2801^{\frac{1}{2}}}{14},\:u=\frac{-55-2801^{\frac{1}{2}}}{14}

Substitute u=x^{\frac{1}{3}} back in:

x^{\frac{1}{3}}=\frac{-55+2801^{\frac{1}{2}}}{14}

\left(x^{\frac{1}{3}}\right)^3

=x^{\frac{1}{3}\cdot \:3}

=x

\left(\frac{-55+2801^{\frac{1}{2}}}{14}\right)^3

=\frac{\left(-55+2801^{\frac{1}{2}}\right)^3}{14^3}

=\left(-55\right)^3+3\left(-55\right)^2\cdot \:2801^{\frac{1}{2}}+3\left(-55\right)\left(2801^{\frac{1}{2}}\right)^2+\left(2801^{\frac{1}{2}}\right)^3

=11876\cdot \:2801^{\frac{1}{2}}-628540

=\frac{11876\cdot \:2801^{\frac{1}{2}}-628540}{14^3}

11876\cdot \:2801^{\frac{1}{2}}-628540\\

=4\left(2969\cdot \:2801^{\frac{1}{2}}-157135\right)

=\frac{4\left(2969\cdot \:2801^{\frac{1}{2}}-157135\right)}{14^3}

=\frac{2^2\left(2969\cdot \:2801^{\frac{1}{2}}-157135\right)}{2^3\cdot \:7^3}

=\frac{2969\cdot \:2801^{\frac{1}{2}}-157135}{7^3\cdot \:2^{3-2}}

=\frac{2969\cdot \:2801^{\frac{1}{2}}-157135}{7^3\cdot \:2}

=\frac{2969\cdot \:2801^{\frac{1}{2}}-157135}{7^3\cdot \:2}

=\frac{2969\cdot \:2801^{\frac{1}{2}}}{686}-\frac{157135}{686}

x=\frac{2969\cdot \:2801^{\frac{1}{2}}}{686}-\frac{157135}{686}

Now solve x^{\frac{1}{3}}=\frac{-55-2801^{\frac{1}{2}}}{14}:

x^{\frac{1}{3}}=\frac{-55-2801^{\frac{1}{2}}}{14}

\left(x^{\frac{1}{3}}\right)^3=\left(\frac{-55-2801^{\frac{1}{2}}}{14}\right)^3

\left(x^{\frac{1}{3}}\right)^3

=x^{\frac{1}{3}\cdot \:3}

=x

\left(\frac{-55-2801^{\frac{1}{2}}}{14}\right)^3

=\frac{\left(-55-2801^{\frac{1}{2}}\right)^3}{14^3}

\left(-55-2801^{\frac{1}{2}}\right)^3

=\left(-55\right)^3-3\left(-55\right)^2\cdot \:2801^{\frac{1}{2}}+3\left(-55\right)\left(2801^{\frac{1}{2}}\right)^2-\left(2801^{\frac{1}{2}}\right)^3

=-628540-11876\cdot \:2801^{\frac{1}{2}}

=\frac{\left(-628540-11876\cdot \:2801^{\frac{1}{2}}\right)}{14^3}

=\frac{-628540-11876\cdot \:2801^{\frac{1}{2}}}{14^3}

-628540-11876\cdot \:2801^{\frac{1}{2}}

=-4\left(157135+2969\cdot \:2801^{\frac{1}{2}}\right)

=-\frac{4\left(157135+2969\cdot \:2801^{\frac{1}{2}}\right)}{14^3}

=-\frac{2^2\left(157135+2969\cdot \:2801^{\frac{1}{2}}\right)}{2^3\cdot \:7^3}

=-\frac{157135+2969\cdot \:2801^{\frac{1}{2}}}{7^3\cdot \:2^{3-2}}

=-\frac{157135+2969\cdot \:2801^{\frac{1}{2}}}{7^3\cdot \:2}

=-\frac{157135+2969\cdot \:2801^{\frac{1}{2}}}{686}

=-\left(\frac{157135}{686}\right)-\left(\frac{2969\cdot \:2801^{\frac{1}{2}}}{686}\right)

=-\frac{157135}{686}-\frac{2969\cdot \:2801^{\frac{1}{2}}}{686}

x=-\frac{157135}{686}-\frac{2969\cdot \:2801^{\frac{1}{2}}}{686}

Solutions:

x=\frac{2969\cdot \:2801^{\frac{1}{2}}}{686}-\frac{157135}{686},\:x=-\frac{157135}{686}-\frac{2969\cdot \:2801^{\frac{1}{2}}}{686}

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7 0
4 years ago
The volume of this cylinder is 2,769.48 cubic meters, the height is 18 cubic meters. What is the radius? Use ​ ≈ 3.14 and round
Firdavs [7]

Answer: r = 7 meters

Step-by-step explanation:

The formula for determining the volume of a cylinder is expressed as

Volume = πr²h

Where

r represents the radius of the cylinder.

h represents the height of the cylinder.

π is a constant whose value is 3.14

From the information given,

Volume = 2,769.48 cubic meters

Height = 18 meters

Therefore,

2769.48 = 3.14 × r² × 18

2769.48 = 56.52r²

r² = 2769.48/56.52 = 49

Taking square root of the left hand side and the right hand side of the equation, it becomes

r = √ 49 = 7 meters

5 0
3 years ago
Best Buy decreased the cost of a Sony flat screen monitor from $525 to $430. What is the percent
valina [46]
The answer would be 1.2
8 0
3 years ago
A quantity with an initial value of 6200 decays continuously at a rate of 5.5% per month. What is the value of the quantity afte
ELEN [110]

Answer:

410.32

Step-by-step explanation:

Given that the initial quantity, Q= 6200

Decay rate, r = 5.5% per month

So, the value of quantity after 1 month, q_1 = Q- r \times Q

q_1 = Q(1-r)\cdots(i)

The value of quantity after 2 months, q_2 = q_1- r \times q_1

q_2 = q_1(1-r)

From equation (i)

q_2=Q(1-r)(1-r)  \\\\q_2=Q(1-r)^2\cdots(ii)

The value of quantity after 3 months, q_3 = q_2- r \times q_2

q_3 = q_2(1-r)

From equation (ii)

q_3=Q(1-r)^2(1-r)

q_3=Q(1-r)^3

Similarly, the value of quantity after n months,

q_n= Q(1- r)^n

As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,

q_{48}=Q(1-r)^{48}

Putting Q=6200 and r=5.5%=0.055, we have

q_{48}=6200(1-0.055)^{48} \\\\q_{48}=410.32

Hence, the value of quantity after 4 years is 410.32.

4 0
3 years ago
Read 2 more answers
Find the indicated length and explain your reasoning
koban [17]

Answer:

LM=25

MN=12

Now,

(M एउटै विन्दु हो)

LM+MN= LN

25+12=37

4 0
2 years ago
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