All categories

3 years ago

3+(-2)-(-3)-(5)+(-1)-(-7)=

Mathematics

1 answer:

Lerok [7]3 years ago

6 0

Answer:

-1

Step-by-step explanation:

did math on caculator

You might be interested in

Need help fast as possible!!!! Pleaseeeee!!!!!!!!

nordsb [41]

ita 3 fiths x plus three eigths

6 0

2 years ago

I need help on problem number 12 and 14 plzzzz

Eva8 [605]

c and a are your answer

7 0

3 years ago

The number of students who attended Albany Medical College last year was 1000. This year, attendance has increased to 1350 stude

NeTakaya

First you have to subtract 1350-1000 to get 350. Then you have to make 350/1000 into a percent. First you have to turn it into a decimal. In the fraction 350/1000 can be reduced to 35/100. In decimal form it would be 0.35. In pecentage form it would be 35%. The percentage difference of the student population would be 35%.

4 0

3 years ago

write the logarithm as a sum or difference of logarithms. simplify each term as much as possible. log4(6ab)

Dominik [7]

The logarithm written as a sum of logarithm and simplified as much as possible is $\frac{1}{2} +log_{4 }3 + log_{4 }a + log_{4 }b$

<h3>Simplifying Logarithms</h3>

From the question, we are to write the given logarithm expression as a sum or difference of logarithms

The given logarithm is

$log_{4 }6ab$

This can be written as

$log_{4 }6 \times a \times b$

From one of the rules of logarithm, we have that

$log_{x }yz= log_{x }y + log_{x }z$

Thus,

$log_{4 }6 \times a \times b$ becomes

$log_{4 }6 + log_{4 }a + log_{4 }b$

This can be further simplified into

$log_{4 }3 \times 2 + log_{4 }a + log_{4 }b$

$log_{4 }3 + log_{4 }2 + log_{4 }a + log_{4 }b$

If desired, this can be further simplified into

$log_{4 }3 + log_{2^{2} }2 + log_{4 }a + log_{4 }b$

$log_{4 }3 + \frac{1}{2} log_{2}2 + log_{4 }a + log_{4 }b$

$log_{4 }3 + \frac{1}{2} (1)+ log_{4 }a + log_{4 }b$

$\frac{1}{2} +log_{4 }3 + log_{4 }a + log_{4 }b$

Hence, the logarithm written as a sum of logarithm and simplified as much as possible is $\frac{1}{2} +log_{4 }3 + log_{4 }a + log_{4 }b$

Learn more on Simplifying logarithms here: brainly.com/question/17851187

#SPJ1

8 0

1 year ago

Two cards are selected from a standard deck of 52 playing cards. The first card is not replaced before the second card is select

Blizzard [7]

Answer:

Probability is: $\frac{\textbf{13}}{\textbf{51}}$

Step-by-step explanation:

From a deck of 52 cards there are 26 black cards. (Spades and Clubs).

Also, there are 26 red cards. (Hearts and Diamonds).

First, we determine the probability of drawing a black card.

P(drawing a black card) = $\frac{number \hspace{1mm} of \hspace{1mm} black \hspace{1mm} cards}{total \hspace{1mm} number \hspace{1mm} of \hspace{1mm} cards}$ $= \frac{26}{52} = \frac{\textbf{1}}{\textbf{2}}$

Now, since we don't replace the drawn card, there are only 51 cards.

But the number of red cards is still 26,

∴ P(drawing a red card) = $\frac{number \hspace{1mm} of \hspace{1mm} red \hspace{1mm} cards}{total \hspace{1mm} number \hspace{1mm}of \hspace{1mm} cards}$ $= \frac{26}{51}$

Now, the probability of both black and red card = $\frac{1}{2} \times \frac{26}{51}$

$= \frac{\textbf{13}}{\textbf{51}}$

Hence, the answer.

5 0

3 years ago