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Nookie1986 [14]
3 years ago
9

Find the value of the expression 3x for the following values of x when x = 10

Mathematics
1 answer:
ElenaW [278]3 years ago
3 0

Answer:

A) 30

Step-by-step explanation:

3*10=30

A 30

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Solve equation: –6 + 3x = –9
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3x=-9+6
3x=-3
x=-3/3
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What is the area of a circle with a radius of 35 inches<br> in2<br> (Use 3.14 for Pi.)
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Answer: area= 3846.5 inch^2

Step-by-step explanation:

Area of circle = πr^2

Where r is the radius and pi is 3.14

So, we get..

3.14 x (35)^2

= 3.14 x 35 x 35

= 3.14 x 1225

= 3846.5 inch^ 2

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3 years ago
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Let <img src="https://tex.z-dn.net/?f=i" id="TexFormula1" title="i" alt="i" align="absmiddle" class="latex-formula"> be the imag
VLD [36.1K]

Hey~freckledspots!\\----------------------

We~will~solve~for~i^{425}!

Rule~of~exponent: a^{b + c} = a^ba^c\\Apply:~i^{425}~=~i^{424}i\\ \\Rule~of~exponent: a^{bc} = (a^{b})^c\\Apply: i^{424} = i(i^2)^{212} \\\\Rule~of~imaginary~number: i^2 = -1\\Apply: i(i^2)^{212} = -1^{212}i\\\\Rule~of~exponent~if~n~is~even: -a^n = a^n\\Apply: -1^{212}i = 1^{212}i\\\\Simplify: 1^{212}i = 1i\\Multiply: 1i * 1 = i\\----------------------\\

Now~let's~solve~1^{14}!\\\\Rule~of~exponent: a^{b + c} = a^ba^c\\Apply: i^{14} = (i^2)^7\\\\Rule~of~imaginary~number: i^2 = -1\\Apply: (i^2)^7 = -1^7\\\\Rule~of~exponent~if~n~is~odd: (-a)^n = -a^n\\Apply: -1^7 = -1^7\\\\Simplify: -1^7 = -1\\----------------------\\Now,~we~have: i-1+i^{-14}+i^{44}\\----------------------

Now~lets~solve~i^{-14}\\\\Rule~of~exponent: a^{-b} = \frac{1}{a^b} \\Apply: i^{-14} = \frac{1}{i^{14}} \\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: \frac{1}{i^{14}} = \frac{1}{(i^2)^7}\\ \\Rule~of~imagianry~number: i^2 = -1\\Apply: \frac{1}{(i^2)^7} = \frac{1}{-1^7} \\\\Simplify: \frac{1}{-1^7} = \frac{1}{-1} \\\\Rule~of~fractions: \frac{a}{-b} = -\frac{a}{b} \\Apply: \frac{1}{-1} = -\frac{1}{1} = -1\\----------------------\\Now,~we~have: i-1-1+i^44\\----------------------

Now~let's~solve~i^{44}!\\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: i^{44} = (i^2)^{22}\\\\Rule~of~imaginary~numbers: i^2 = -1\\Apply: (i^2)^{22} = -1^{22}\\\\Rule~of~exponent~if~n~is~even: (-a)^n = a^n\\Apply: -1^{22} = 1^{22}\\\\Simplify: 1^{22} = 1\\----------------------\\Now,~we~have~i-1-1+1\\----------------------

Now~let's~simplify~the~expression!\\\\= i-1-1+1 \\= 1 + i -2\\= -1+i\\----------------------

Answer:\\\large\boxed{-1+i}\\----------------------

Hope~This~Helped!~Good~Luck!

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3 years ago
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Michael puts $800 into a savings account. The account earns 2% simple interest. How much interest will he earn in 6 months?
expeople1 [14]

Answer:

he earned 8 dollars in interest

Step-by-step explanation:

800(1+(.02)(.5))

hopefully im correct

8 0
2 years ago
Consider randomly selecting a student at a certain university, and let A denote the event that the selected individual has a Vis
Mice21 [21]

Answer:

Step-by-step explanation:

Given that,

Visa card is represented by P(A)

MasterCard is represented by P(B)

P(A)= 0.6

P(A')=0.4

P(B)=0.5

P(B')=0.5

P(A∩B)=0.35

1. P(A U B) =?

P(A U B)= P(A)+P(B)-P(A ∩ B)

P(A U B)=0.6+0.5-0.35

P(A U B)= 0.75

The probability of student that has least one of the cards is 0.75

2. Probability of the neither of the student have the card is given as

P(A U B)'=1-P(A U B)

P(A U B)= 1-0.75

P(A U B)= 0.25

3. Probability of Visa card only,

P(A)= 0.6

P(A) only means students who has visa card but not MasterCard.

P(A) only= P(A) - P(A ∩ B)

P(A) only=0.6-0.35

P(A) only=0.25.

4. Compute the following

a. A ∩ B'

b. A ∪ B'

c. A' ∪ B'

d. A' ∩ B'

e. A' ∩ B

a. A ∩ B'

P(A∩ B') implies that the probability of A without B i.e probability of A only and it has been obtain in question 3.

P(A ∩ B')= P(A-B)=P(A)-P(A∩ B)

P(A∩ B')= 0.6-0.35

P(A∩ B')= 0.25

b. P(A ∪ B')

P(A ∪ B')= P(A)+P(B')-P(A ∩ B')

P(A ∪ B')= 0.6+0.5-0.25

P(A ∪ B')= 0.85

c. P(A' ∪ B')= P(A')+P(B')-P(A' ∩ B')

But using Demorgan theorem

P(A∩B)'=P(A' ∪ B')

P(A∩B)'=1-P(A∩B)

P(A∩B)'=1-0.35

P(A∩B)'=0.65

Then, P(A∩B)'=P(A' ∪ B')= 0.65

d. P( A' ∩ B' )

Using demorgan theorem

P(A U B)'= P(A' ∩ B')

P(A U B)'= 1-P(A U B)

P(A' ∩ B')= 1-0.75

P(A' ∩ B')= 0.25

P(A U B)'= P(A' ∩ B')=0.25

e. P(A' ∩ B)= P(B ∩ A') commutative law

Then, P(B ∩ A') = P(B) only

P(B ∩ A') = P(B) -P(A ∩ B)

P(B ∩ A') =0.5 -0.35

P(B ∩ A') =0.15

P(A' ∩ B)= P(B ∩ A') =0.15

5 0
2 years ago
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