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Goshia [24]
3 years ago
8

Write a story problem to represent 3÷5=3/5

Mathematics
1 answer:
DIA [1.3K]3 years ago
7 0
3÷5=3/5
3/5=0,6
3÷5=0,6
You might be interested in
Find the value of X<br><br>A: 2<br>B: 4<br>C: 6<br>D: 8​
11111nata11111 [884]

Using the Pythagorean theorem:

x^2 + (x-2)^2 = (√20)^2

Simplify the right side:

x^2 + (x-2)^2 = 20

Subtract 20 from both sides:

x^2 + (x-2)^2 - 20 = 0

Factor:

(x-4)(x+2) = 0

Solve for each x:

x = 4 and x = -2

The side cant be a negative value, so the answer would be x = 4

The answer is B.

3 0
3 years ago
Help Plz!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Ivan
Place your dot roughly around 2.23.  The square root of 5 is 2.23
5 0
3 years ago
Describe how to determine the average rate of change between x = 3 and x = 5 for the function f(x) = 3x3 + 2. Include the averag
Serhud [2]

Answer:

147

Step-by-step explanation:

Average rate of change is given by the total rise over total run.

f(x) = 3x^3+2\\f(3)=3*3^3+2=83\\f(5)=3*5^3+2 = 377

So,

total rise = 377 - 83 = 294

total run = 2

average rate of change = \dfrac{294}{2} = 147

8 0
2 years ago
Will mark as brainliest! 40 points! Thanks!
n200080 [17]

When given a system of equations, the "solutions" are defined where two equations intersect, or meet.


A. The point where the lines p(x) and g(x) meet is (3, -1), and thus this is considered the solution set.

B. Because there are three lines in total, g(x) is able to intersect both lines one time, and so it has two pairs of solutions.

The first is (3, -1), which has already been established with p(x).

The second is (0, 5), and this is where it intersects with f(x).

C. The solution to f(x) = g(x) is 0, as this is the only x value where both equations are equal.


Hope my answer helped!

6 0
3 years ago
Read 2 more answers
Which are the solutions of x^2 = –11x + 4?
strojnjashka [21]

This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get

x^2 + 11x - 4 = 0

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2}

where x_{1,2} is a compact way to indicate both solutions x_1 and x_2, while a,b,c are the coefficients of the quadratic equation, i.e. we consider the polynomial ax^2+bx+c.

So, in your case, we have a=1,\ \ b=11,\ \ c=-4

Plug those values into the formula to get

x_{1,2} = \frac{-11\pm\sqrt{121+16}}{2} = \frac{-11\pm\sqrt{137}}{2}

So, the two solutions are

x_1 = \frac{-11+\sqrt{137}}{2}

x_2 = \frac{-11-\sqrt{137}}{2}

3 0
3 years ago
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