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zzz [600]
3 years ago
7

Oliver and Peter had the

Mathematics
2 answers:
lana66690 [7]3 years ago
6 0
They both used the same amount. 6/8 when simplified is the same as 3/4.
Law Incorporation [45]3 years ago
4 0

Answer:

Oliver and Peter both uses same length string.

Explanation:

Suppose length of rope or string = x meters

Oliver divide his rope into four equal parts

\frac{1}{4},  \frac{1}{4}, \frac{1}{4}, \frac{1}{4}

Each part length is \frac{1}{4}

NOw Oliver used 3 part of the string out of four parts

that means total used string is 3/4

Now peter also have same length of string = x meter

He divides the string eight equal parts and use the 6 parts out of 8

each part length is :\frac{x}{8} , \frac{x}{8} , \frac{x}{8} , \frac{x}{8} , \frac{x}{8} , \frac{x}{8} , \frac{x}{8} , \frac{x}{8}

Use 6x/8

simplify the fraction 6x/8

divide by 2 both numerator and denominator

6x/8 = 3x / 4

So both uses same part of string.

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98 is what percent of 280?
joja [24]

Answer:

35%

Step-by-step explanation:

Find the percent by dividing 98 by 280

98/280

= 0.35

So, 98 is 35% of 280

3 0
3 years ago
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Which statement is true about the following system of equations
GenaCL600 [577]
\left\{\begin{array}{ccc}\frac{3x-1}{2y}=1\\x+2y=7\end{array}\right\\\\\left\{\begin{array}{ccc}\frac{3x-1}{2y}=1\\2y=7-x\end{array}\right\\\\substitute:\\\\\frac{3x-1}{7-x}=1\iff3x-1=7-x\\\\3x+x=7+1\\4x=8\\x=2\\\\2y=7-2\\2y=5\\y=2.5\\\\Answer:D)\ y > x
6 0
4 years ago
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Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte
Mashcka [7]

Answer:

f'(x) > 0 on (-\infty ,\frac{1}{e}) and f'(x)<0 on(\frac{1}{e},\infty)

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

f'(x)>0

To find its decreasing interval :

f'(x)

2) Then let's find the critical point of this function:

f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0

2.2 Solving for x this equation, this will lead us to one critical point since x'  is not defined for Real set, and x''=\frac{1}{e}≈0.37 for e≈2.72

x^{2x}=0 \ni \mathbb{R}\\(ln(x)+1)=0\Rightarrow ln(x)=-1\Rightarrow x=e^{-1}\Rightarrow x\approx 2.72^{-1}x\approx 0.37

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.

8 0
3 years ago
What’s the answer to this? You have to estimate
bazaltina [42]
0.19579831 (that's what my math app said the estimate was)
4 0
4 years ago
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#3 please find f (5)
DerKrebs [107]

Answer:

f(5)=1/22

Step-by-step explanation:

f(5)=-2/-5(5)-19

f(5)=-2/-25-19

f(5)=-2/-44

f(5)=1/22

8 0
3 years ago
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