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denis-greek [22]
3 years ago
11

Show the difference of the two sets

Mathematics
1 answer:
Rudik [331]3 years ago
6 0
There is 8 fruits in set A and 7 in set B
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The vertices of a triangle are formed by the intersections of the following three lines:
irakobra [83]

Answer:

y=x

Step-by-step explanation:

y=12

7 0
4 years ago
Find the bases for Col A and Nul​ A, and then state the dimension of these subspaces for the matrix A and an echelon form of A b
Rainbow [258]

Answer:

skip counting by 0

Step-by-step explanation:

skipcount by 0 to get to 100 for the third column.

3 0
4 years ago
Which of the following best describes a circle
Marrrta [24]

Answer:

The definition of a circle is the set of all points at a given distance (it's radius) from a given point (it's center)

Step-by-step explanation:

5 0
3 years ago
In the coordinate plane, choose the graph with the conditions given.<br><br> x = -3
finlep [7]

You have not given any graphs to choose from but x=-3 would be the graph that has a vertical line running through every point that has -3 as an x coordinate and would cross the x axis (the horizontal axis) at -3

5 0
3 years ago
Using descartes' rule of signs to describe the roots of h(x)=4x^4-5x^3+2x^2-x+5, how many total roots must there be in this four
vlabodo [156]
H(x) = 4x⁴ - 5x³ + 2x² - x + 5

Take the coefficients, we have:

4   -5   2   -1   5

From the first coefficient to the second coefficient, there's a change of sign from positive to negative (positive four to negative five)

From the second coefficient to the third coefficient, there's a change of sign from negative to positive (negative five to positive two)

From the third coefficient to the fourth coefficient, there's a change of sign from positive to negative (positive two to negative one)

From the fourth coefficient to the fifth coefficient, there's a change of sign from negative to positive (negative one to positive five)

The changing of sign happened four times, and it means h(x) has four positive real zeros or less (even numbers of zeros), so h(x) could have either four or two real zeros.

The maximum number of solutions for a polynomial of degree four is four solution, so there are no negative zeros. 
4 0
3 years ago
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