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o-na [289]
3 years ago
9

What is 1.08 to the power of

\frac{1}{5}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
OlgaM077 [116]3 years ago
6 0

Answer:

about \: 1.01551

Step-by-step explanation:

No it isn't, because:

{1.08}^{ \frac{1}{5} }  =  { \frac{108}{100} }^{ \frac{1}{5} }  =  { \frac{27}{25} }^{ \frac{1}{5} }  =  \sqrt[5]{ \frac{27}{25} }  = 1.01551

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Write the equation of a line that is
zvonat [6]

Answer:

y=0.5x-5

Step-by-step explanation:

Since the equation is parallel, this means that the slope remains the same which is 0.5.

The remaining thing to do is to find b or y-intercept. To do that, the point (6,-2) and the slope of 0.5 will be used.

-2=0.5(6)+b

When solving for b, it will equal to -5.

Therefore the new equation that is parallel is y=0.5x-5

6 0
3 years ago
6. Mary worked 27 hours last week. If she gets paid $12/hour, what is her<br> gross pay?
kupik [55]

Answer:

324

Step-by-step explanation:

you would multiply 27 × 12 and that would equal 324

8 0
2 years ago
If 2 sides of a triangle are 6 and 16, what is the range of the possible lengths of the third side?
Leno4ka [110]

Answer:

between 10 and 22

Step-by-step explanation:

A side of a triangle must measure between the difference of the lengths of the other two sides and the sum of the lengths of the other two sides.

Difference: 16 - 6 = 10

Sum: 6 + 16 = 22

Answer: between 10 and 22

6 0
3 years ago
the sum of 49,000 naira is to be shared among A, B, and C in the ratio of 1:2:4, respectively. find how much each wilk receive.​
Natalka [10]

Answer:

#7000, #14000 and #28000 respectively

Step-by-step explanation:

#49000 to be shared in 1:2:4

49000 ÷ (1 + 2 + 4) = 49000/7 = #7000

A will get (7000×1) = #7000

B will get (7000×2) = #14000

C will get (7000×4) = #28000

3 0
3 years ago
How many distinct pairs of disjoint non-empty subsets of A are there, the union of which is all of A?
Mrac [35]
A = {1, 2, 5, 6, 8}
{1} U {2, 5, 6, 8} 
{2} U {1, 5, 6, 8} 
{5} U {1, 2, 6, 8} 
{6} U {1, 2, 5, 8} 
{8} U {1, 2, 5, 6} 
{1, 2} U {5, 6, 8} 
{1, 5} U {2, 6, 8} 
{1, 6} U {2, 5, 8}
{1, 8} U {2, 5, 6} 
{1, 2, 5} U {6, 8} 
{1, 2, 6} U {5, 8}
{1, 2, 8} U {5, 6}
{1, 5, 6} U {2, 8}
{1, 5, 8} U {2, 6}
{1, 6, 8} U {2, 5} 
The answer is 15 distinct pairs of disjoint non-empty subsets.
5 0
3 years ago
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