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fenix001 [56]
3 years ago
5

REALLY NEED HELP TONIGHT!!!!!

Mathematics
1 answer:
AfilCa [17]3 years ago
4 0
<span>(5a</span>⁴<span>b</span>²<span>)</span>³<span>(-2b</span>⁴<span>) =  (125a</span>¹²b⁶<span>)(-2b</span>⁴<span>) = -250a</span>¹²b¹⁰
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oee [108]

Answer:

The answer is True. Just trust me on this one.

8 0
2 years ago
Need help with this. Hope someone can help
Triss [41]
Hi,

Let me help you with this problem.

v = ( \frac{2 \times 1.5}{2} ) \times 3 \\ v = \frac{3}{2} \times 3 \\ v = 1.5 \times 3 \\ v = 4.5

Answer: 4.5m3

Hope this helps.
r3t40
7 0
3 years ago
Can someone please give me the answers to this? ... please ...
Darina [25.2K]

Step-by-step explanation:

anyway, without any further information about ground consistency, friction and stuff, we have to assume ideal circumstances.

so, the same energy that made the ball flying through the air makes it also bounce or roll across the ground carrying it exactly the same distance at the same time.

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6 0
3 years ago
What is the number when 0.83 is written as a fraction with a denominator of 100?
Grace [21]
The fraction would be 83/100
7 0
3 years ago
Read 2 more answers
PLEASE HELPP ♥
kirza4 [7]

Answer:

An exponential function is a function of the form

f(x)=bx

where b≠1 is a positive real number. The domain of an exponential function is (−∞,∞) and the range is (0,∞).

Solve the equation: 52x−3=752x−3=7.

Since we can’t easily rewrite both sides as exponentials with the same base, we’ll use logarithms instead. Above we said that logb(x)=ylogb⁡(x)=y means that by=xby=x. That statement means that each exponential equation has an equivalent logarithmic form and vice-versa. We’ll convert to a logarithmic equation and solve from there.

52x−3log

⎛⎝⎜

⎞⎠⎟=7=2x−352x−3=7log5

⁡(7

)=2x−3

From here, we can solve for xx directly.

2xx=log5(7)+3=log5(7)+32

A logarithmic function is a function defined as follows

logb(x)=ymeans thatby=xlogb⁡(x)=ymeans thatby=x

where b≠1b≠1 is a positive real number. The domain of a logarithmic function is (0,∞)(0,∞) and the range is (−∞,∞)(−∞,∞).

Solve the equation:

log3(2x+1)=1−log3(x+2).log3⁡(2x+1)=1−log3⁡(x+2).

With more than one logarithm, we’ll typically try to use the properties of logarithms to combine them into a single term.

log3(2x+1)log3(2x+1)+log3(x+2)log3((2x+1)(x+2))log3(2x2+5x+2)2x2+5x+22x2+5x−1=1−log3(x+2)=1=1=1=3=0log3⁡(2x+1)=1−log3⁡(x+2)log3⁡(2x+1)+log3⁡(x+2)=1log3⁡((2x+1)(x+2))=1log3⁡(2x2+5x+2)=12x2+5x+2=32x2+5x−1=0

Let’s use quadratic formula to solve this.

x=−5±52−4⋅2⋅−1−−−−−−−−−−−√2⋅2=−5±

−−−−−−−−⎷4.x=−5±52−4⋅2⋅−12⋅2=−5±33

4.

What happens if we try to plug x=

4 0
3 years ago
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