All categories

3 years ago

Will give brainliest! thanks :)

Mathematics

2 answers:

jek_recluse [69]3 years ago

4 0

Answer: A

Step-by-step explanation:

Soloha48 [4]3 years ago

4 0

Answer:

A is the answer

Step-by-step explanation:

Congruent means same size and the same shape, but they may not be the same orientation

You might be interested in

Suppose m = 2 + 6i, and | m + n | = 3√10, where n is a complex number.

Ksju [112]

Answer: a) √50

b) n = 1 + 7i

Step-by-step explanation:

first, the modulus of a complex number z = a + bi is

IzI = √(a^2 + b^2)

The fact that n is complex does not mean that n doesn't has a real part, so we must write our numbers as:

m = 2 + 6i

n = a + bi

Im + nI = 3√10

Im + n I = √(a^2 + b^2 + 2^2 + 6^2)= 3√10

= √(a^2 + b^2 + 40) = 3√10

a^2 + b^2 + 40 = 3^2*10 = 9*10 = 90

a^2 + b^2 = 90 - 40 = 50

√(a^2 + b^2 ) = InI = √50

The modulus of n must be equal to the square root of 50.

now we can find any values a and b such a^2 + b^2 = 50.

for example, a = 1 and b = 7

1^2 + 7^2 = 1 + 49 = 50

Then a possible value for n is:

n = 1 + 7i

6 0

3 years ago

asmine wants to use her savings of $1128 to buy video games and music CDs. The total price of the music CDs she bought was $72.

Inga [223]

Let g represent video games. The equation you can use to solve for the amount of video games and CDs Jasmine can buy with her savings is:
43g+72=1128

The 43g part is because it costs 43 dollars for every video game, so you need to multiply 43 by the amount of games she buys to get the total for g games.
The +72 part is because you already have the amount spent on CDs, and needs to be added to the cost of video games, so that the sum would be equal to 1128.

Now, solving:
43g+72=1128 ...subtract 72 from both sides
43g=1056 ...divide both sides by 43 to isolate g
g=24.5581395349
Rounded up, that would be 25 games, but Jasmine's savings isn't enough to buy 25 games that cost $43 each, so you would go down to 24 games.

The answer: Jasmine can buy 24 games with her savings after spending $72 on CDs.

8 0

3 years ago

Help plsss!?!?!??!?!??>...................................

Soloha48 [4]

Answer:

A,B,D

Step-by-step explanation:

case A) If the price is marked down by percent, the new price will be

we know that

If the price is marked down by 10 percent

then the new price will be

10%= 10/100=0.10

(1 - 0.10) *$28= 0.90 * $28

=$25.20

therefore The statement case A) is True

case B) If the price is marked down by 25 percent, the new price will be $21

we know that

If the price is marked down by 25 percent

then the new price will be

25%=25/100=0.25

(1-0.25)* $28=0.75*$28

=$21

therefore The statement case B) is True

case C) If the price is marked down by 50 percent, the new price will be $19 we know that

If the price is marked down by 50 percent

then the new price will be

50%=50/100=0.50

(1-0.50)*$28= 0.50*$28

=14

therefore The statement case C) is False

case D) If the price is marked down by 35 percent, the new price will be $18.20

we know that

If the price is marked down by 35 percent

then the new price will be

35%= 35/100=0.35

(1-0.35)* $28=0.65*$28

=18.20

therefore The statement case D) is True

case E) If the price is marked down by 40 percent, the new price will be $29.80 we know that

If the price is marked down by 40 percent

then the new price will be

40%=40/100=0.40

(1-0.40)*$28= 0.60*$28

=$16.80

therefore The statement case E) is False

7 0

2 years ago

Suppose that you pick a bit string from the set of all bit strings of length ten. Find the probability that a) the bit string ha

emmasim [6.3K]

Answer:

45/1024
1/4
15/128
193/512
9/512

Step-by-step explanation:

There are 2^10 = 1024 bit strings of length 10.

a) There are 10C2 = 45 ways to have exactly two 1-bits in 10 bits

p(2 1-bits) = 45/1024

b) Of the four (4) possibilities for beginning and ending bits (00, 01, 11, 10), exactly one (1) of those is 00.

p(b0=0 & b9=0) = 1/4

c) There are 10C7 = 120 ways to have seven 1-bits in the bit string.

p(7 1-bits) = 120/1024 = 15/128

d) ∑10Ck {for k=0 to 4} = 386 is the total of the number of ways to have 0, 1, 2, 3, or 4 1-bits in the string. If there are more than that, there won't be more 0-bits than 1-bits

p(more 0 bits) = 386/1024 = 193/512

e) The string will have two 1-bits if it starts with 1 and there is a single 1-bit among the other 9 bits. There are 9 ways that can happen, among the 512 ways to have 9 remaining bits.

p(2 1-bits | first is a 1-bit) = 9/512

6 0

3 years ago

In shipment of 850 widgets, 32 are found to be detective. At this rate, how many detective widgets could be expected in 22,000 w

cluponka [151]

Answer:

The number of detective widgets out of 22,000 widgets is 828 widgets

Step-by-step explanation:

Given as :

The total number of widgets = 850

The number of detective widgets out of 850 = 32

Now,

∵ Out of 850 widgets , the number of detective widgets = 32

SO,Out of 1 widgets , the number of detective widgets = $\frac{32}{850}$

∴ Out of 22,000 widgets , the number of detective widgets = $\frac{32}{850}$ × 22,000

Or, Out of 22,000 widgets , the number of detective widgets = 828.23

Hence, The number of detective widgets out of 22,000 widgets is 828 widgets Answer

4 0

3 years ago