Answer:
<em> </em><em>-</em><em>1</em>
Step-by-step explanation:
<em>here's</em><em> your</em><em> solution</em>
<em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>The </em><em>value</em><em> of</em><em> x</em><em> </em><em>=</em><em> </em><em>1</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>The </em><em>value</em><em> of</em><em> </em><em>y </em><em>=</em><em> </em><em>2</em>
<em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>we </em><em>need</em><em> to</em><em> find</em><em> </em><em>(</em><em> </em><em>x </em><em>-</em><em> </em><em>y)</em>
<em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>so,</em><em> </em><em>put </em><em>the </em><em>value </em><em>of </em><em>above</em><em> </em><em>variables</em><em> </em><em>in </em><em>it</em>
<em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>hence</em><em>,</em><em> </em><em>(</em><em>1</em><em> </em><em>-</em><em> </em><em>2</em><em>)</em><em> </em><em>=</em><em> </em><em>-</em><em> </em><em>1</em>
<em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>-</em><em> </em><em>1</em><em> </em><em>is </em><em>correct</em><em> answer</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em>hope</em><em> it</em><em> helps</em>
Answer:
This problem is incomplete, we do not know the fraction of the students that have a dog and also have a cat. Suppose we write the problem as:
"In Mrs.Hu's classroom, 4/5 of the students have a dog as a pet. X of the students who have a dog as a pet also have cat as a pet. If there are 45 students in her class, how many have both a dog and a cat as pets?"
Where X must be a positive number smaller than one, now we can solve it:
we know that in the class we have 45 students, and 4/5 of those students have dogs, so the number of students that have a dog as a pet is:
N = 45*(4/5) = 36
And we know that X of those 36 students also have a cat, so the number of students that have a dog and a cat is:
M = 36*X
now, we do not have, suppose that the value of X is 1/2 ("1/2 of the students who have a dog also have a cat")
M = 36*(1/2) = 18
So you can replace the value of X in the equation and find the number of students that have a dog and a cat as pets.
The value of y is 3 when x is 6.
Solution:
The given table is the amount of cheese in ounces and the cost price.
Let x represents the amount of cheese and y represents the cost price.
The formula which derived from the table is y = x ÷ 2.
That is
.
<u>To find the value of y when x is 6:</u>

Substitute x = 6 in the formula, we get

y = 3
Hence the value of y is 3 when x is 6.
Answer:
Jordan will have $ 910.00 after 3 years
Step-by-step explanation
Answer would be C because the cost is independent