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Dominik [7]
3 years ago
6

Given the sets above, select all (or the null set) of the element of: B

Mathematics
1 answer:
Elanso [62]3 years ago
5 0

Given:

Universal set = {a, b, c, d, e, f, g, h}

A = {a, b, c, d}

B = {a, d, e}

C= {c, f, g, h}

To find:

The elements of set B'.

Solution:

Let universal set be U.

U = {a, b, c, d, e, f, g, h}

We know that,

Compliment of a set = Universal set - Set

It means,

B'=U-B

B'=\{a, b, c, d, e, f, g, h\}-\{a,d,e\}

B'=\{b, c,f, g, h\}

Therefore, only b, c, f, g, h are the elements of set B'.

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Tema [17]

6⁴ means "multiply 6 by 6 4 times"


Or



6·6·6·6

5 0
3 years ago
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guapka [62]

Answer:

Step-by-step explanation:

Use Pythagorean theorem

Base² + altitude² = hypotenuse²

Base² + 7² = 20²

Base² + 49 = 400

Base² = 400 - 49 = 351

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5 0
2 years ago
The 5th out of 10 questions
seraphim [82]

In the division a/b, a is the dividend, and b is the divisor.

In this problem, 2/3 is the dividend, and 4/5 is the divisor.

She needed to multiply the dividend, 2/3, by the reciprocal of the divisor, 5/4.

Instead, she multiplied the dividend by the divisor without using the reciprocal.

Answer: second choice

7 0
3 years ago
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2) Winston needs 80 signatures from students in his school before he can run for class president. He has 23
Marizza181 [45]

Answer: 19

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80 - 23 = 57 signatures left

57 signatures divided by him and his 2 friends

57 / 3 = 19

They need 19 signatures each

7 0
3 years ago
Which type of triangle is formed with the points A(1, 7), B(-2, 2), and C(4, 2) as its vertices?
Dafna1 [17]
We will have to use the distance formula in order to determine the lengths of each side of the triangle.

Distance formula: \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} }

Let's calculate AB first:
A (1, 7) and B (-2, 2)
A: x1 = 1 and y1 = 7
B: x2 = -2 and y2 = 2

so
\sqrt{(-2 - 1)^{2} + (2 - 7)^{2} }
\sqrt{(-3)^{2} + (-5)^{2} }
\sqrt{9 + 25 }
AB = \sqrt{34} or (rounded to the nearest tenth) ≈ 5.8

Now let's do BC:
B: x1 = -2 and y1 = 2
C: x2 = 4 and y2 = 2

So
\sqrt{(4 - -2)^{2} + (2 - 2)^{2} }
\sqrt{(6)^{2} + (0)^{2} }
BC = \sqrt{36 } or 6

Now let's do CA
C: x1 = 4 and y1 = 2
A: x2 = 1 and y2 = 7

So
\sqrt{(1 - 4)^{2} + (7 - 2)^{2} }
\sqrt{(-3)^{2} + (5)^{2} }
\sqrt{9 + 25}
CA = \sqrt{34} or (rounded to the nearest tenth) ≈ 5.8

So let's recap:

AB ≈ 5.8
BC = 6
CA ≈ 5.8

So AB and AC are the same length while BC is .2 units longer which means this is an isosceles triangle.

4 0
3 years ago
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