The measure of 1 in the diagram below is 113º. What is the measure of <4?

Find a counterexample to show that the statement is false. Assume all sets are subsets of a universal set U = {1, 2, 3, 4, 5}. (

natka813 [3]

Answer:

For subsets A={1,2},B={2,3},C={1,2 5} of U = {1, 2, 3, 4, 5}.

A ∪ (B − C) = (A ∪ B) − (A ∪ C) does not hold.

Step-By-Step Explanation:

U = {1, 2, 3, 4, 5}.

A={1,2}

B={2,3}

C={1,2 5}

For the Left Hand Side

B-C={3}

A ∪ (B − C)= {1,2} ∪ {3} ={1,2,3}

Likewise, the Right Hand Side:

(A ∪ B)={1,2,3}

(A ∪ C)={1,2,5}

(A ∪ B)- (A ∪ C)= {1,2,3}-{1,2,5}={3}

Since, {1,2,3} ≠ {3}, A ∪ (B − C) = (A ∪ B) − (A ∪ C) does not hold.

7 0

3 years ago

In circle S with the measure of angle RST = 102 and RS = 6 units, find the length of arc RT. Round to the nearest hundredth

GREYUIT [131]

Answer:

10.68

Step-by-step explanation:

\text{Arc Length}=

Arc Length=

\,\,\frac{\text{Central Angle}}{360}(\text{Circumference})

360

Central Angle

(Circumference)

RT=

\,\,\frac{102}{360}(2 \pi6)

360

102

(2π6)

RT=

\,\,10.68

10.68

3 0

3 years ago

Joseph is creating a garden up against his fence in his backyard in the shape of a semi-circle. The diameter of his garden is 8

Marizza181 [45]

Answer:

25 inches of soil

Step-by-step explanation:

7 0

3 years ago

If the discriminant is negative then the quadratic has

zepelin [54]

Answer:

No real solutions (or has two complex roots)

Step-by-step explanation:

The discriminant, b² - 4ac, is the expression (radicand ) of the quadratic equation:

$x = \frac{-b +/- \sqrt{b^{2}-4ac} }{2a}$ .

The value of the discriminant determines the nature and the number of solutions given by the quadratic equation.

A discriminant with a negative value (or b² - 4ac < 0 ) means that the quadratic equation has no real solutions or two complex solutions. Quadratic equations with a negative discriminant also have no x-intercepts; thus, its graph will not cross the x-axis.

4 0

3 years ago

Factor completely 3x ^2 -27

Tom [10]

First, factor out a 3.
3(x² - 9)

In any quadratic ax² + bx + c, we can split the bx term up into two new terms which we want to equal the product of a and c.
In this case, we have x² + 0x - 9. (the 0x is a placeholder)
We want two numbers that add to 0 and multiply to get -9.
Obviously, these numbers are 3 and -3.

Now we have 3(x² + 3x - 3x - 9).
Let's factor.
3(x(x+3)-3(x+3))
3(x-3)(x+3)

There are multiple shortcuts which you could make here, FYI:
Instead of splitting the middle, if your a value is 1, you can go straight to that step (x+number)(x+other number).
Whenever you have a difference of squares, like a²-b², that factors to (a+b)(a-b).

3 0

3 years ago

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