All categories

3 years ago

The Difference between two numbers is 22. The sum of the same two numbers is 78. What are the two numbers?

Mathematics

1 answer:

jenyasd209 [6]3 years ago

3 0

Answer:

50 and 28

Step-by-step explanation:

x+y=78

x-y=22

to find this, subtract 22 from 78 to get 56

divide 56 by 2 to get 28

y=28

add 22 to 28 to get 50

x=50

to double check add 50 and 28 to get 78

also subtract 28 from 50 to get 22

You might be interested in

Will charges $12.50 per hour for painting plus a flat $15 fee for preparation and clean-up. Which linear equation shows the amou

Y_Kistochka [10]

Is the answer W sorry if it is wrong

5 0

3 years ago

Imran made a profit of 20 percent

Ivanshal [37]

He overall had a 10% loss.

6 0

2 years ago

To create a histogram which set of intervals would be the most appropriate to split the following data set? 120, 133, 134, 125,

Schach [20]

Put them in order then find the median and if there are two or more divided by how many are in the middle

8 0

3 years ago

Describe the characteristics of each set of numbers that make up the set of integers

Greeley [361]

Numbers I think I'm not that smart

8 0

2 years ago

15 pts. Prove that the function f from R to (0, oo) is bijective if - f(x)=x2 if r- Hint: each piece of the function helps to "c

solniwko [45]

Answer with explanation:

Given the function f from R to $(0,\infty)$

f: $R\rightarrow(0,\infty)$

$-f(x)=x^2$

To prove that the function is objective from R to $(0,\infty)$

Proof:

$f:(0,\infty )\rightarrow(0,\infty)$

When we prove the function is bijective then we proves that function is one-one and onto.

First we prove that function is one-one

Let $f(x_1)=f(x_2)$

$(x_1)^2=(x_2)^2$

Cancel power on both side then we get

$x_1=x_2$

Hence, the function is one-one on domain [tex[(0,\infty)[/tex].

Now , we prove that function is onto function.

Let - f(x)=y

Then we get $y=x^2$

$x=\sqrt y$

The value of y is taken from $(0,\infty)$

Therefore, we can find pre image for every value of y.

Hence, the function is onto function on domain $(0,\infty)$

Therefore, the given $f:R\rightarrow(0.\infty)$ is bijective function on $(0,\infty)$ not on whole domain R .

Hence, proved.

3 0

3 years ago