All categories

2 years ago

Find the values of x and y. 2(5x- 5)° (5y + 5)º (7y - 9° (6x + 50)

Mathematics

1 answer:

Amanda [17]2 years ago

4 0

The values of x and y are 15 and 7 respectively

The angles formed from the intersection of two lines at X are equal and vertically opposite angles.

From the figure shown, we can conclude that:

2(5x-5) = 6x + 50 (vertically opposite angle)

Expand

10x - 10 = 6x + 50

10x - 6x = 50 + 10

4x = 60

x = 60/4

x =15

Similarly;

5y + 5 = 7y - 9

Collect the like terms;

5y - 7y = -9 - 5

-2y = -14

y = 14/2

y = 7

Hence the values of x and y are 15 and 7 respectively

Learn more here: brainly.com/question/15198257

You might be interested in

One problem with the widespread use of email is that you receive too much of it.

Kobotan [32]

The answer for your question is True because it says that "Widespread" by that you already know widespread means that it goes every where and you would get it more then one so for that your answer is true:)

Hope this helps:)

6 0

3 years ago

How many solutions does the equation −2a + 2a + 7 = 8 have? One Two None Infinitely many

aleksandrvk [35]

Answer:

no solutions

Step-by-step explanation:

−2a + 2a + 7 = 8

Combine like terms

7 = 8

This is never true so there are no solutions

5 0

3 years ago

Read 2 more answers

The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

Anon25 [30]

We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

If we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

Let us know find its inverse.

$y=(x-5)^{2}+2$

Upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

Let us now solve this function for y.

$(y-5)^{2}=x-2\\
y-5=\pm \sqrt{x-2}\\
y=5\pm \sqrt{x-2}\\$

Hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .