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Novay_Z [31]
3 years ago
9

THREE QUESTIONS CONCERNING QUADRATIC EQUATIONS: (a) State the four methods used to solve quadratic equations (b) Name the four t

ypes of solution types used to classify quadratic equations particularly when the discriminant is used. (c) The discriminant of a quadratic equation is -25. What type of solution(s) does the equation have?
Mathematics
1 answer:
alexira [117]3 years ago
6 0

Answer:

A. Factoring; Completing the Square; Quadratic Formula; Graphing

B. All real numbers; 2 complex roots; 1 real root; 2 distinct real roots

C. 2 complex roots

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4×7<br> 4×7=4(_+_)<br> 4×7=(4×_)+(4×_)<br> 4×7=_+_<br> 4×7=_
Lady bird [3.3K]

Answer:

4+3 =7 then 7×4=28

hoped it helped I am not sure if it is right

5 0
3 years ago
Read 2 more answers
April invested in a lifetime annuity that begins payments at age 65. Her life expectancy is 89. April invested a $1,000,000 into
Mila [183]

Answer:

Monthly payment is $5970

Step-by-step explanation:

Given : April invested in a lifetime annuity that begins payments at age 65. Her life expectancy is 89. April invested a $1,000,000 into the annuity, which earns 4.4% APR compounded monthly.

To find : what is April's monthly payment amount?

Solution :

Formula of monthly payment  

Monthly payment, M=\frac{\text{Amount}}{\text{Discount factor}}  

Discount factor D=\frac{1-(1+i)^{-n}}{i}  

Where, Amount = $1,000,000

Rate r= 4.4%=0.044 compounded monthly

i=\frac{0.044}{12}=0.004  

Time = 89-65=23 years

n=23\times12=276  

Now, put all the values we get,  

D=\frac{1-(1+i)^{-n}}{i}  

D=\frac{1-(1+0.004)^{-276}}{0.004}  

D=\frac{1-(1.004)^{-276}}{0.004}  

D=\frac{1-0.33}{0.004}  

D=\frac{0.67}{0.004}  

D=167.5  

Monthly payment, M=\frac{\text{Amount}}{\text{Discount factor}}  

M=\frac{1000000}{167.5}  

M=5970.1  

Approximately,monthly payment is $5970

4 0
4 years ago
Read 2 more answers
Im bored lemme get yallz IG's please dont report :) oh happy saint patricks day :)
nikklg [1K]

Answer:7+7

Step-by-step explanation:

i didn't know it was st.patrick's day

4 0
3 years ago
What is the GCF of 100xyz and 25xz
kvv77 [185]

Answer:

25

Step-by-step explanation:

25 goes into 100 4 times, 25 goes into 25 once  i think not sure

6 0
3 years ago
1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff
Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If  A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that  f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0
4 years ago
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