80 x 90/100 = 72
Therefore there were 72 milkshakes with whipped cream that were sold.
Answer:
37
Step-by-step explanation:
There are 2 ways to do this, the first is to just count the number of shaded squares (35)
The second is to find the area of the larger shaded square and subtract the smaller un-shaded rectangles
The large square is 7x7=49 squares
There are two 2x3 rectangles, which are 6 each
49-6-6=37
Rectangle perimeter is 53 meters.
L = W + 2.3
To find the options of a perimeter, divide by 4.
= 13.25
So if this was a square, it would be 13.25 for both L and W but it's not.
Since the W has 2.3 meters more then the L, we do this.
13.25 + 2.3 = 15.55
13.25 - 2.3 = 10.95
Two of your sides equal 15.55, and the other two equal 10.95.
Hope this helps!
Answer:
Choice b.
.
Step-by-step explanation:
The highest power of the variable
in this polynomial is
. In other words, this polynomial is quadratic.
It is thus possible to apply the quadratic formula to find the "roots" of this polynomial. (A root of a polynomial is a value of the variable that would set the polynomial to
.)
After finding these roots, it would be possible to factorize this polynomial using the Factor Theorem.
Apply the quadratic formula to find the two roots that would set this quadratic polynomial to
. The discriminant of this polynomial is
.
.
Similarly:
.
By the Factor Theorem, if
is a root of a polynomial, then
would be a factor of that polynomial. Note the minus sign between
and
.
- The root
corresponds to the factor
, which simplifies to
. - The root
corresponds to the factor
, which simplifies to
.
Verify that
indeed expands to the original polynomial:
.
Answer: x=4, -1
Step-by-step explanation:
Assuming you meant
, the zeros of the question are x = 4 and -1.
Step 1. Replace f(x) with y.

Step 2. To find the roots of the equation, replace <em>y</em> with 0 and solve.

Step 3. Factor the left side of the equation.

Step 4. Set x-4 equal to 0 and solve for <em>x</em>.

Step 5. Set
equal to 0 and solve for <em>x</em>.

The solution is the result of
and
.
