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3 years ago

What are the values of x and y in the equation

Mathematics

1 answer:

vaieri [72.5K]3 years ago

5 0

Answer:

The correct choice is D

Step-by-step explanation:

Value of X=20 and then to know the value of y we need to add

52+52= 104

360-104= 256

256÷2= 128

Y=128

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Step-by-step explanation:

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3 years ago

Please help me with this question

Anettt [7]

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Read 2 more answers

One number is times a first number. A third number is 100 more than the first number. If the sum of the three numbers

icang [17]

When a number is 3 times a first number. A third number is 100 more than the first number and the sum of the three numbers is 450, the numbers are 70, 170 and 210.

<h3>How to calculate the numbers?</h3>

Let the first number = x

Second number = 3x

Third number = x + 100

Sum = 450

The numbers will be:

x + 3x + x + 100 = 450

5x + 100 = 450

5x = 450 - 100

5x = 350

Divide

x = 350 / 5

x = 70

First number = 70

Second number = 3x = 3 × 70 = 210

Third number = x + 100 = 70 + 100 = 170

Therefore, the numbers are 70, 170 and 210.

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brainly.com/question/24644930

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One number is 3 times a first number. A third number is 100 more than the first number. If the sum of the three numbers is 450 , find the numbers.

8 0

1 year ago

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

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3 years ago

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nata0808 [166]

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3 years ago