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saw5 [17]
3 years ago
5

Find 20% of 120. Show work please.

Mathematics
2 answers:
icang [17]3 years ago
7 0

Answer:

24

Step-by-step explanation:

When you are finding a percent of something, you need to move the decimal over to the left twice in the percentage. Like this:

20% = .20

Then, just multiply that by the number you are trying to find the percent of.

.20 x 120 = 24

Triss [41]3 years ago
7 0

Answer: 24

Step-by-step explanation: To find 20% of 120, first write 20% as a decimal by moving the decimal point 2 places to the left to get <em>.</em>20.

Next, the word "of" means multiply, so we multiply <em>.</em>20 by 120.

(.20) (120) = 24

Therefore, 20% of 120 is 24.

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Plz help and show work plz
MrRa [10]

Answer:

3rd option

Step-by-step explanation:

The equation of a parabola in vertex form is

f(x) = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Here (h, k ) = (- 1, - 25 ) , then

f(x) = (x - (- 1) )² - 25 , that is

     = (x + 1)² - 25 ← expand using FOIL

     = x² + 2x + 1 - 25

     = x² + 2x - 24

7 0
3 years ago
Write an equation for the line described in standard form with x-intercept 4 and y-intercept 5
stealth61 [152]

Answer:

Equation( \frac{5}{4}) x  + y = 5 is in the standard dorm.

Step-by-step explanation:

Here, the x - intercept = 4 , so ( 4,0) is a point on the line.

and the y - intercept = 5, so ( 0,5) is a point on the line.

So, the slope of the equation is given as  =  \frac{y_2 - y_1}{x_2-x_1 }  = \frac{5 -0}{0-4}  = -\frac{5}{4}

Now, the SLOPE INTERCEPT FORM of an equation is :

y  = mx + b: here m  = slope and b =  y- intercept

or, y =-( \frac{5}{4}) x + 5

Now, standard form is Ax +By = C

So, the standard form is ( \frac{5}{4}) x  + y = 5

Hence, the above equation ( \frac{5}{4}) x  + y = 5

is in the standard dorm.

4 0
3 years ago
Can anyone help me with this
Sedbober [7]

Answer:

x=0, - 9

Step-by-step explanation:

x^2+9x=0

x(x+9)=0

x=0 or x=-9

8 0
2 years ago
Read 2 more answers
At most, how many unique roots will a third-degree polynomial function have?
beks73 [17]

The fundamental theorem of algebra states that a polynomial with degree n has at most n solutions. The "at most" depends on the fact that the solutions might not all be real number.

In fact, if you use complex number, then a polynomial with degree n has exactly n roots.

So, in particular, a third-degree polynomial can have at most 3 roots.

In fact, in general, if the polynomial p(x) has solutions x_1,\ x_2,\ldots x_n, then you can factor it as

p(x) = (x-x_1)(x-x_2)\ldots (x-x_n)

So, a third-degree polynomial can't have 4 (or more) solutions, because otherwise you could write it as

p(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)

But this is a fourth-degree polynomial.

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3 years ago
Help please!!!!!!!!!!!!!!!! Asap
Hunter-Best [27]
One positive and one negative
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3 years ago
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