When a number is decreased by 66%, the result is 13. What is the original number to the nearest tenth?

How do you simplify i²⁹? (If you can't tell, that's the imaginary number i^29).

maria [59]

The imaginary number ' i ' is the square root of -1.

i = √-1
i² = -1
i³ = -√-1
i to the 4th power = +1
i to the 5th power = √-1
etc.

and all the higher powers keep going in the same 4-step cycle.

What's 29 divided by 4 ?
It's 7 with a remainder of 1.

So 29 with all the 4s thrown away is 1, and ' i ' to the 29th power is
the same thing as ' i ' to the 1st power = √-1 or just plain ' i '.

6 0

2 years ago

Algebra. Find the variable in the equation 803 divided by 73=m

weeeeeb [17]

<u>Answer:</u>

The value of variable "m" in the equation 803 divided by 73 = m is equal to 11.

<u>Solution:</u>

Need to find the variable in the given equation 803 divided by 73 = m

That means need to solve below expression to determine value of variable m.

$\frac{803}{73}=m$

So let’s solve left hand side to get the value of m.

For sake of simplicity let’s do factorization of numerator and denominator and remove the common terms in numerator and denominator.

$\begin{array}{l}{803=73 \times 11} \\\\ {73=73 \times 1} \\\\ {=>\frac{803}{73}=\frac{73 \times 11}{73 \times 1}=11}\end{array}$

Hence value of variable m is equal to 11.

8 0

2 years ago

The perimeter of a pool is 150 m. the rectangle at the right is a scale drawing of the pool. The length of each square on the gr

Anon25 [30]

Answer:

Step-by-step explanation:

From the given rectangle,

Length of the rectangle = 5 cm

Width of the rectangle = 10 cm

Ratio of length : width = 1 : 2

Perimeter of the rectangle = 2(l + b)= 150 cm

l + b = 75 -----(1)

Since, l : b = 1 : 2

$\frac{l}{b}= \frac{1}{2}$

b = 2l ------(2)

From equations (1) and (2),

l + 2l = 75

3l = 75

l = 25 m

Width 'b' = 2(25) = 50 m

Now we have to draw a rectangle with a scale = 25 : 2

Therefore, Scale factor = $\frac{25}{l}=\frac{25}{2}$

l = 2 cm

Similarly,

$\frac{50}{w}=\frac{25}{2}$

w = 4 cm

Now we can draw a rectangle with length = 2 cm and width = 4 cm.

5 0

2 years ago

Jillian is selling boxes of cookies to raise money for her basketball team. The 10 oz. box costs $3.50, while the 16 oz. box cos

RideAnS [48]

Answer:

Jillian sold 15 ten-oz bags and 9 sixteen-oz bags.

Step-by-step explanation:

Let x represent number of 10-oz bags and y represent number of 16-oz bags.

We have been given that Jillian sold 24 boxes of cookies in total. We can represent this information in an equation as:

$x+y=24...(1)$

We are also told that 10 oz. box costs $3.50, while the 16 oz. box costs $5.00, so the cost of x 10-oz bags would be $3.50x$ and cost of y 16-oz bags would be $5y$ .

At the end of one week, she collected $97.50. We can represent this information in an equation as:

$3.5x+5y=97.50...(2)$

Now, we will use substitution method to solve our system of equations. From equation (1), we will get:

$y=24-x$

Substitute this value in equation (2):

$3.5x+5(24-x)=97.50$

$3.5x+120-5x=97.50$

$-1.5x+120=97.50$

$-1.5x+120-120=97.50-120$

$-1.5x=-22.5$

$\frac{-1.5x}{-1.5}=\frac{-22.5}{-1.5}$

$x=15$

Therefore, the number of 10-oz boxes sold be Jillian is 15.

Now, we will substitute $x=15$ in equation (1) to solve for y.

$15+y=24$

$15-15+y=24-15$

$y=9$

Therefore, the number of 16-oz boxes sold be Jillian is 9.

8 0

2 years ago

Jane is taking two courses. The probablity she passes the first course is 0,7. The probablity she passes the second course is 0.

makkiz [27]

Answer:

a. 0.6

b. not independent

c. 0.1

d. 0.4

e. 0.3

Step-by-step explanation:

a.

P(passing first course)=P(C1)=0.7

P(passing second course)=P(C2)=0.8

P(passing at least one course)=P(C1∪C2)=0.9

P( passes both courses)=P(C1∩C2)=?

We know that

P(A∪B)=P(A)+P(B)-P(A∩B)

P(A∩B)=P(A)+P(B)-P(A∪B)

So,

P( passes both courses)=P(C1∩C2)=P(C1)+P(C2)-P(C1∪C2)

P( passes both courses)=P(C1∩C2)=0.7+0.8-0.9

P( passes both courses)=P(C1∩C2)=0.6

Thus, the probability she passes both courses is 0.6.

b.

The event of passing one course is independent of passing another course if

P(C1∩C2)=P(C1)*P(C2)

P(C1)*P(C2)=0.7*0.8=0.56

P(C1∩C2)=0.6

As,

0.6≠0.56

P(C1∩C2)≠P(C1)*P(C2),

So, the event of passing one course is dependent of passing another course.

c.

P(not passing either course)=P(C1∪C2)'=1-P(C1∪C2)

P(not passing either course)=P(C1∪C2)'=1-0.9

P(not passing either course)=P(C1∪C2)'=0.1

Thus, the probability of not passing either course is 0.1.

d.

P(not passing both courses)=P(C1∩C2)'=1-P(C1∩C2)

P(not passing both courses)=P(C1∩C2)'=1-0.6

P(not passing both courses)=P(C1∩C2)'=0.4

Thus, the probability of not passing both courses is 0.4.

e.

P(passing exactly one course)=?

P(passing exactly course 1)=P(C1)-P(C1∩C2)=0.7-0.6=0.1

P(passing exactly course 2)=P(C2)-P(C1∩C2)=0.8-0.6=0.2

P(passing exactly one course)=P(passing exactly course 1)+P(passing exactly course 2)

P(passing exactly one course)=0.1+0.2

P(passing exactly one course)=0.3

Thus, the probability of passing exactly one course is 0.3.

3 0

2 years ago