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kirill115 [55]
3 years ago
7

14. Set A contains all integers from 50 to 100, inclusive, and Set B contains all integers from 69 to 138, exclusive. How many i

ntegers are included in both Set A and Set B?
E. 29
F. 30
G. 31
H. 32​
Mathematics
1 answer:
tamaranim1 [39]3 years ago
5 0

Answer:

30

Step-by-step explanation:

We have to make the intersection between these sets, and we see that they intersect in the range: [70,100], and d(70,100) = 30, that is the answer.

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A set of equations is given below:
nexus9112 [7]

I would think this is many solutions, the reason is because, there is no specific answer (y) so, x could be anything because it would have to equal to y, which could also be anything! I hope im right! Sorry if im wrong!

3 0
3 years ago
One of our brainliest, Konrad509, made this:
Reil [10]

\dfrac{B_x \sqrt{74_x}}{1D_x}+J_x51_x=4G3_x

A=10, B=11, C=12, etc.

\dfrac{11\cdot x^0\cdot \sqrt{7\cdot x^1+4\cdot x^0}}{1\cdot x^1+13\cdot x^0}+19\cdot x^0\cdot (5\cdot x^1+1\cdot x^0)=4\cdot x^2+16\cdot x^1+3\cdot x^0\\\\\dfrac{11\sqrt{7x+4}}{x+13}+19(5x+1)=4x^2+16x+3\\\\\dfrac{11\sqrt{7x+4}}{x+13}+95x+19=4x^2+16x+3\\\\11\sqrt{7x+4}+95x(x+13)+19(x+13)=(4x^2+16x+3)(x+13)\\\\11\sqrt{7x+4}+95x^2+1235x+19x+247=4x^3+52x^2+16x^2+208x+3x+39\\\\11\sqrt{7x+4}=4x^3-27x^2-1043x-208\\\\121(7x+4)=(4x^3-27x^2-1043x-208)^2

121(7x+4)=(4x^3-27x^2-1043x-208)^2\\\\847x+484=16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433888 x + 43264\\\\16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433041 x +42780=0

Now, the "only" thing that remains to do is solving the above equation.

While making this problem I only made sure it has a solution. I didn't try to solve it myself and I didn't know it will end up with such "convoluted" polynomial. Sorry to everyone who tried to solve it... m(_ _)m

I think the best way to approach it is using the rational root theorem since we know that x\in\mathbb{N}. Moreover we can deduce that x\geq19 since there is J and J=19.

After you succesfully solve it, you should get the answer x=20.

7 0
3 years ago
Can you figure out the formula
puteri [66]

If "a" and "b" are two values of x-coordinate, and "m" is the midpoint between them, it means the distance from one end to the midpoint is the same as the distance from the midpoint to the other end

... a-m = m-b

When we add m+b to this equation, we get

... a+b = 2m

Solving for m gives

... m = (a+b)/2

This applies to y-coordinates as well. So ...

... The midpoint between (x1, y1) and (x2, y2) is ((x1+x2)/2, (y1+y2)/2)

_____

Jennifer had (x1, y1) = (-4, 10) and (x2, y2) = (-2, 6). So her calculation would be

... midpoint = ((-4-2)/2, (10+6)/2) = (-6/2, 16/2) = (-3, 8)

Brandon had (x1, y1) = (9, -4) and (x2, y2) = (-12, 8). So his calculation would be

... midpoint = ((9-12)/2, (-4+8)/2) = (-3/2, 4/2) = (-1.5, 2)

4 0
3 years ago
Please hurry i need help
kirza4 [7]

Answer:

(x^2+16x+63)

Step-by-step explanation:

(x^2+17x+72)-(1x+9)

(x^2+16x+63)

If this is incorrect I appologize

6 0
3 years ago
Read 2 more answers
Does each set of side lengths form a right triangle 10 m, 11 m, 15 m or 15 m, 8 m, 17 m,
IrinaVladis [17]
Can you be a little more specific please
6 0
3 years ago
Read 2 more answers
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