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sesenic [268]
3 years ago
9

6TH GRADE HELP!! Describe the similarities and differences between primary and secondary sources. In two sentences.

Mathematics
2 answers:
denis-greek [22]3 years ago
5 0
A primary source saw something first-hand. A secondary source only heard about it.
Nuetrik [128]3 years ago
3 0
The most basic similarity between primary & secondary sources is "They both Convey Information","Neither is an Absolute Objective Source" and in some special situation, a single source can be both at the same time.

Now, Difference is that, Secondary sources analyzes, interprets and explains primary sources, while they are independent and doesn't include any picture, articles, tools etc. which secondary sources do

Hope this helps!
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My cousin needs help on these questions
antiseptic1488 [7]

Answer:

Number 7.) triangle

Number 8.) Cylinder

Number 3.) Rectangle

Step-by-step explanation:

4 0
3 years ago
A store sells small notebooks for ​$8 and large notebooks for ​$10. if a student buys 6 notebooks and spends ​$56​, how many of
exis [7]
The first thing you have to do is find a way to make the last digit 6 (since its $56). So that would be $8×2=16, so there are 2 small notebooks. Then, she needs another $40 to get $56, so 40÷10=4, so she bought 4 large notebooks.

Answer: 2 small, 4 large
3 0
4 years ago
Read 2 more answers
1. On the last 5 quizzes, Marco has earned a total of 21 out of 25 possible points. If he has a perfect
Nonamiya [84]

The number of points Marco needs to bring his average up to 90% is 15points.

If there's a possible 25 possible points, and Marco has taken 5 quizzes;

Evidently, each quiz has weighs 5 points.

Mathematically;

\frac{21 + 5x}{5(5 + x)}  \times 100\% =  \frac{90}{100}

  • 2100 + 500x = 2250 + 450x

  • 50x = 150

x = 3.

Therefore, he needs 5x points = 15 points to bring his average up to 90%.

Read more:

https://brainly.in/question/24249798

3 0
3 years ago
Nickel-63 is a radioactive substance with a half-life of about 100 years. An artifact had
Marina CMI [18]

Answer:

see below

Step-by-step explanation:

The equation for half life is

n = no  e ^ (-kt)

Where no is the initial amount of a substance , k is the constant of decay and t is the time

no = 9.8

1/2 of that amount is 4.9 so n = 4.9 and t = 100 years

4.9 = 9.8 e^ (-k 100)

Divide each side by 9.8

1/2 = e ^ -100k

Take the natural log of each side

ln(1/2) = ln(e^(-100k))

ln(1/2) = -100k

Divide each side by -100

-ln(.5)/100 = k

Our equation in years is

n = 9.8  e ^ (ln.5)/100 t)

Approximating ln(.5)/100 =-.006931472

n = 9.8 e^(-.006931472 t)  when t is in years

Now changing to days

100 years = 100*365 days/year

36500 days

Substituting this in for t

4.9 = 9.8 e^ (-k 36500)

Take the natural log of each side

ln(1/2) = ln(e^(-36500k))

ln(1/2) = -36500k

Divide each side by -100

-ln(.5)/36500 = k

Our equation in years is

n = 9.8  e ^ (ln.5)/36500 d)

Approximating ln(.5)/365=-.00001899

n = 9.8 e^(-.00001899 d)  when d is in days

3 0
4 years ago
Read 2 more answers
1. Subtract the linear expressions.
jeyben [28]

(1)\\\\(45x -5) -(13x +4) = 45x-5-13x -4 = 32x -9\\\\\\(2)\\\\-2(4n) -3 + (-7)(8n) +2 = -8n-56n -1 = -64n -1 = -(64n+1)\\\\(3)\\\\(-3.6x+9.4) - (1.9x +3.6) = -3.6x +9.4 -1.9x -3.6 =-5.5x+5.8\\\\\\(4)\\\\-3x+4-(5x-2) = -3x+4-5x +2=-8x +6

7 0
3 years ago
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