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3 years ago

Find m<1 if <1 is complementary to <2, <2 is supplementary to <3, and

Mathematics

1 answer:

kirza4 [7]3 years ago

6 0

Answer:

∠ 1 = 36°

Step-by-step explanation:

Supplementary angles sum to 180° then

∠ 2 + ∠ 3 = 180° , that is

∠ 2 + 126° = 180° ( subtract 126° from both sides )

∠ 2 = 54°

Complementary angles sum to 90° , then

∠ 1 + ∠ 2 = 90° , that is

∠ 1 + 54° = 90° ( subtract 54° from both sides )

∠ 1 = 36°

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What is the nth term of 4,9,16,25,36

konstantin123 [22]

Answer:

(n + 1)²

Step-by-step explanation:

note that 4, 9, 16, 25, 36 are square numbers

To generate the terms add 1 to the position n and square, that is

first term : n = 1 → n = 1 + 1 → 2² = 4

second term : n = 2 → n = 2 + 1 → 3² = 9

third term : n = 3 → n = 3 + 1 → 4² = 16

Hence n th term is (n + 1)²

4 0

3 years ago

Let f be a functions of degree 4 whose coefficients are real numbers: two of its zeros are - 3 and 4 - i. Explain why one of the

kvasek [131]

Answer:

Step-by-step explanation:

We have the following theorem, if f is a polynomial with real coefficients, we can factor it completely in factors of the degree at most 2.

Consider first a polynomial of degree two, hence it is a polynomial of the form $ax^2+bx+c$ . The cuadratic formula tells us that the solutions are of the form

$x = \frac{-b\pm \sqrt[]{b^2-4ac}}{2a}$ .

Note that square root, over the reals, tells us that the are only real solutions if $b^2-4ac \geq 0$ . If that is not the case, say it's negative, the solution are complex. Then, the solutions are of the form

$x = \frac{-b \pm i \sqrt[]{4ac-b^2}}{2a}$ . NOte that this means that if we have a complex number of the form a+bi that is a solution, then the number a-bi (who is called the complex conjugate) is also a solution.

Recall that when we have a polynomial f(x) whose a zero is the number c, then we can factor f as follows f(x) = (x-c) * p(x) where p(x) is another polynomial of lesser degree .

So far, we know that -3 and 4-i are zeros of the function f. Note that we are missing two zeros. But, since complex numbers are zeros of polynomial only by pairs (that is the number and its conjugate are zeros), then, we must have that one of the missing 2 zeros is a real number. We have 4-i as a zero, then, its complex conjugate must be also a zero, i.e 4+i is a zero.

8 0

3 years ago

Find the number of elements in A 1 ∪ A 2 ∪ A 3 if there are 200 elements in A 1 , 1000 in A 2 , and 5, 000 in A 3 if (a) A 1 ⊆ A

lina2011 [118]

Answer:

a. 4600

b. 6200

c. 6193

Step-by-step explanation:

Let $n(A)$ the number of elements in A.

Remember, the number of elements in $A_1 \cup A_2 \cup A_3$ satisfies

$n(A_1 \cup A_2 \cup A_3)=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)$

Then,

a) If $A_1\subseteq A_2, n(A_1 \cap A_2)=n(A_1)=200$ , and if $A_2\subseteq A_3, n(A_2\cap A_3)=n(A_2)=1000$

Since $A_1\subseteq A_2\; and \; A_2\subseteq A_3, \; then \; A_1\cap A_2 \cap A_3= A_1$

$n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-200-200-1000-200=4600$

b) Since the sets are pairwise disjoint

$n(A_1 \cup A_2 \cup A_3)=\\n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\200+1000+5000-0-0-0-0=6200$

c) Since there are two elements in common to each pair of sets and one element in all three sets, then

$n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-2-2-2-1=6193$

8 0

3 years ago

Wha are the equations. of the asymptotes of the graph of the function f(x)=3x^2-2x-1/x^2+3x+10

Leviafan [203]

The asymptote is the line the graph comes close to but never touches, so if you graph it out you should be able to see it! Sorry I couldn't help any more but good luck

3 0

3 years ago

Mr.Sampson fills the family pool with water for the summer, after two hours, the water has reached the depth of 2.5 feet.After t

scoray [572]

This question is incomplete

Complete Question

Mr Sampson fills the family pool with water for the summer.After two hours,the water has reached a depth of 2.5 feet. After three hours,the water level has risen to 3 3/4 feet.If the relationship between time and water depth is proportional,what is the constant of proportionality?

Answer:

1.33

Step-by-step explanation:

Time is proportional to Depth

Hence:

T ∝ D

T = kD

Where k is the constant of proportionality

After two hours,the water has reached a depth of 2.5 feet. After three hours,the water level has risen to 3 3/4 feet.

Total number of hours now is 3 + 2 = 5 hours

5 = k × 3 3/4

5 = k × 3.75

k = 5/3.75

k = 1.33

8 0

3 years ago