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2 years ago

What is the solution of the system? 10x – 2y = 24 6x+2y = 8

Mathematics

1 answer:

Stels [109]2 years ago

7 0

Answer: x = 2 , y = 5

Step-by-step explanation:

10x - 2y = 24 ..................... equation 1

6x + 2y = 8 ........................ equation 2

solving the system of linear equation by elimination method , add equation 1 and 2

16x = 32

divide through by 16

x = 2

substitute x = 2 into equation 1 to find the value of y

10(2) - 2y = 2y

20 - 2y = 2y

20 = 4y

y = 5

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Solve the equation: k^2+5k+13=0

mr_godi [17]

Step-by-step explanation:

k² + 5k + 13 = 0

Using the quadratic formula which is

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

From the question

a = 1 , b = 5 , c = 13

So we have

$k = \frac{ - 5 \pm \sqrt{ {5}^{2} - 4(1)(13) } }{2(1)} \\ = \frac{ - 5 \pm \sqrt{25 - 52} }{2} \\ = \frac{ - 5 \pm \sqrt{ - 27} }{2} \: \: \: \: \: \: \\ = \frac{ - 5 \pm3 \sqrt{3} \: i}{2} \: \: \: \: \: \:$

Separate the solutions

$k_1 = \frac{ - 5 + 3 \sqrt{3} \: i }{2} \: \: \: \: or \\ k_2 = \frac{ - 5 - 3 \sqrt{3} \: i}{2}$

The equation has complex roots

Separate the real and imaginary parts

We have the final answer as

$k_1 = - \frac{5}{2} + \frac{3 \sqrt{3} }{2} \: i \: \: \: \: or \\ k_2 = - \frac{5}{2} - \frac{3 \sqrt{3} }{2} \: i$

Hope this helps you

8 0

2 years ago

if you earn $144 in 12 hours how much would it be in 17 hours? (i need in a algebraic equatiionnnnnn)

Studentka2010 [4]

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2 years ago

Find the annual rate of interest. Principal = 4600 rupees, Period = 5 years, Total amount = 6440 rupees, Annual rate of interest

Len [333]

The annual rate of interest per year is 8%

Solution:

Given:- Principal (p) = 4600 rupees, Time –Period (t) = 5 years, Total amount(A) = 6440 rupees

First we will calculate the Interest and then using formula of simple interest we will calculate the rate of interest

Interest = Amount – Principal

Interest = 6440 – 4600 = 1840

Now using the formula of simple Interest and on putting values we get,

$\text {Simple Interest }=\frac{P \times R \times T}{100}$

Where "P" is the principal and "R" is the rate of interest per annum and "T" is the time period

$1840=\frac{4600 \times R \times 5}{100}$

$\begin{aligned} \mathrm{R} &=\frac{1840 \times 100}{4600 \times 5} \\\\ \mathrm{R} &=\frac{1840}{230} \\\\ \mathrm{R} &=8 \% \end{aligned}$

Hence, the required rate of interest per year is 8%

6 0

2 years ago

What is the measure of ∠EGF? __

Ray Of Light [21]

Answer:

∠EGF = 1/2( 180 - 50) = 1/2(130) = 65

∠CGF = 180 - 65 = 115.

Step-by-step explanation:

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Gnesinka [82]

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Step-by-step explanation:

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