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fredd [130]
3 years ago
8

Compare the value of 3 in 302754 and 53891. The value of 3 in 302254 is _______ times the value of the 3 and 53891

Mathematics
1 answer:
noname [10]3 years ago
8 0
Compare how many place values the 3's have.
The 3 in 302754 is in the 6th place value, while the 3 in 53891 is in the 4th place value. Each place value is 10 times the one below it. Since the first 3 is 2 place values ahead, 10*10=100 times.
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Triangle: solve for x
NARA [144]

Answer:

6 root two

Step-by-step explanation:

geometric mean for leg t.

24*3=x2

x2=72

x=root(72)

6 root two

if my answer helps please mark as brainliest.

6 0
3 years ago
Marry Emma earned $45.00 babysitting her cousin. She saved $18.50 and spent the remainder. How much did Mary Emma spend?
gizmo_the_mogwai [7]

Answer:

She spent $26.50

Step-by-step explanation:

subtract $18.50 from $45.00

3 0
3 years ago
Factorize:<br>3a^2 + 10a + 3​
sergey [27]

Answer: (3a + 1) (a + 3)

Step-by-step explanation:

<u>Concept:</u>

Here, we need to know the idea of factorization.

It is like "splitting" an expression into a multiplication of simpler expressions. Factoring is also the opposite of Expanding.

<u>Solve:</u>

Given = 3a² + 10a + 3

<em>STEP ONE: separate 3a² into two terms</em>

3a

a

<em>STEP TWO: separate 3 into two terms</em>

3

1

<em>STEP THREE: match the four terms in ways that when doing cross-multiplication, the result will give us 10a.</em>

3a     1

a       3

When cross multiply, 3a × 3 + 1 × a = 10a

<em>STEP FOUR: combine the expression horizontally to get the final factorized expression.</em>

3a  ⇒   1

a   ⇒    3

(3a + 1) (a + 3)

Hope this helps!! :)

Please let me know if you have any questions

7 0
3 years ago
Write result in standard notation
levacccp [35]

Answer:

7.5 * 10 ^- 4

Step-by-step explanation:

6 0
3 years ago
Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,
posledela

Answer:

  • number of multiplies is n!
  • n=10, 3.6 ms
  • n=15, 21.8 min
  • n=20, 77.09 yr
  • n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

  mpy[n] = n·mp[n-1]

  mpy[2] = 2

So, ...

  mpy[n] = n! . . . n ≥ 2

__

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

  10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

  n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

  n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

  n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0
3 years ago
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