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3 years ago

MATHS QUESTION Any help for another maths questions I just don't get what to do thank you ❤

Mathematics

1 answer:

levacccp [35]3 years ago

4 0

Look at the picture.

a) 180° - 45° = 135°

b) Use trigonometric function

$\tan(\angle DBA)=\dfrac{45}{30}\\\\\tan(\angle DBA)=1.5\\\\m\angle DBA\approx56^o$

270° - 56° = 214°

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Monica paid sales tax of $1.50 when she bought a new bike helmet. If the sales tax rate was 5% how much did the store charge for

Salsk061 [2.6K]

$1.45
1.50 times .05= 0.075
the sales tax was 0.075 cents, so if you subtract that from $1.50, you get the price before sales tax

7 0

3 years ago

Brainly users, do your magic please.

Keith_Richards [23]

Answer:

c.) 8/3

Step-by-step explanation:

the two sides must weigh the same, or 12 grams.

the weights labelled x combined with the 4 grams weight should equal 12 grams.

the equation solved for x would be:

3x + 4 = 12

3x = 8

x = 8/3

6 0

3 years ago

A cryptographic hash takes a message as input and produces a fixed-length string as output, called the digital fingerprint. A br

Allushta [10]

Answer:

$1.85x10^{19}$ attempts are required to find a matching pair if the digital fingerprint is 64 bits long.

$3.40*10^{38}$ attempts are required to find a matching pair if the digital fingerprint is 128 bits long.

Step-by-step explanation:

Each bit has two options. So

How many attempts are required to find a matching pair if the digital fingerprint is 64 bits long?

So for each of the 64 bits, we have the following number of options.

2 - 2 - 2 - 2 -... - 2

So, in all, there are

$T = 2^{64} = 1.85x10^{19}$

options.

So, $1.85x10^{19}$ attempts are required to find a matching pair if the digital fingerprint is 64 bits long.

128 bits long?

Using the same logic as the first question.

$T = 2^{128} = 3.40*10^{38}$

So, $3.40*10^{38}$ attempts are required to find a matching pair if the digital fingerprint is 128 bits long.

5 0

3 years ago

What is the probability of randomly choosing a permutation of the 10 digits 0, 1, 2, . . . , 9 in which (a) an odd digit is in t

inysia [295]

The first digit is one of {1, 3, 5, 7, 9}, and the last digit is one of {1, 2, 3, 4, 5}.

If the first digit is one of {1, 3, 5} (3 choices), then the last digit is one of {1, 2, 3, 4, 5} minus whatever is picked for the first digit (4 choices, and the remaining eight digits can be arranged in 8! ways, so there are 12*8! possible permutations.

If the first digit is one of {7, 9} (2 choices, then the last digit is one of {1, 2, 3, 4, 5} (5 choices), and again there are 8! ways of arranging the remaining digits, so there are 10*8! possible permutations.

Then the total number of permutations that fit the criteria is 12*8! + 10*8! = 22*8!. There are 10! total permutations that can be made overall, so the probability of randomly picking one we want is

$\dfrac{22\cdot8!}{10!}=\dfrac{22}{10\cdot9}=\boxed{\dfrac{11}{45}}$

7 0

4 years ago

138 trash bags cost 16.50 what is the cost of one bag

Vesna [10]

You device 16.50÷130= .12

4 0

3 years ago