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Dmitrij [34]
3 years ago
15

Which equation, when solved, results in a different value of x than the other three?

Mathematics
2 answers:
mamaluj [8]3 years ago
8 0

Answer: d

Step-by-step explanation: got it right on edge

marusya05 [52]3 years ago
6 0
HtvygjnvtYVCTVHNKLPLDDDDTCVTBYTV WHATa
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Angles 1 and 2 are supplementary. 2 lines intersect to form angles 1 and 2. which equation represents the relationship between t
timama [110]

Answer:

angle 1+angle 2 = 180 degrees

Step-by-step explanation:

when 2 angles are supplementary, they equal 180 degrees when they are added up. Because angles 1 and 2 are supplementary, this means that angle 1 + angle 2 equals 180 degrees.

6 0
2 years ago
What is the length of the arc intercepted by an angle of 10 degrees on a circle with a radius of 10 meters?
soldier1979 [14.2K]
   
\displaystyle\\
\texttt{Length of the arc }= 2\pi R\times  \frac{10}{360} = \frac{2\pi \times 10}{36} = \frac{20\pi}{36} =  \boxed{\frac{5\pi}{9}~m}



3 0
3 years ago
Write out the first few terms of the series Summation from n equals 0 to infinity (StartFraction 2 Over 3 Superscript n EndFract
anyanavicka [17]

Answer:

\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = 15/8

Step-by-step explanation:

The sum you are trying to understand is this.

\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n}

Remember that in general when you have a geometric series  

\sum\limits_{n = 0}^{\infty} a*r^n you have that

\sum\limits_{n = 0}^{\infty} a*r^n = \frac{a}{1-r}      and that equality is true as long as     |r| < 1.

Therefore here we have

\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{3*5} \big)^n = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n        and   \big|\frac{-1}{15} \big| = \frac{1}{15} < 1

Therefore we can use the formula and

\sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n =  \frac{2}{1-(-1/15)} = \frac{2}{1+1/15} = 30/16  = 15/8

5 0
3 years ago
A rancher wants to fence in an area of 1500000 square feet in a rectangular field and then divide it in half with a fence down t
VLD [36.1K]

Answer:

6000 ft

Step-by-step explanation:

Let length of rectangular field=x

Breadth of rectangular field=y

Area of rectangular field=1500000 square ft

Area of rectangular field=l\times b

Area of rectangular field=x\time y

1500000=xy

y=\frac{1500000}{x}

Fencing used ,P(x)=x+x+y+y+y=2x+3y

Substitute the value of y

P(x)=2x+3(\frac{1500000}{x})

P(x)=2x+\frac{4500000}{x}

Differentiate w.r.t x

P'(x)=2-\frac{4500000}{x^2}

Using formula:\frac{dx^n}{dx}=nx^{n-1}

P'(x)=0

2-\frac{4500000}{x^2}=0

\frac{4500000}{x^2}=2

x^2=\frac{4500000}{2}=2250000

x=\sqrt{2250000}=1500

It is always positive because length is always positive.

Again differentiate w.r.t x

P''(x)=\frac{9000000}{x^3}

Substitute x=1500

P''(1500)=\frac{9000000}{(1500)^3}>0

Hence, fencing is minimum at x=1 500

Substitute x=1 500

y=\frac{1500000}{1500}=1000

Length of rectangular field=1500 ft

Breadth of rectangular field=1000 ft

Substitute the values

Shortest length of fence used=2(1500)+3(1000)=6000 ft

Hence, the shortest length of fence that the rancher can used=6000 ft

3 0
4 years ago
On Tuesday the temperature is 1 ºC. By Wednesday it has dropped to –2 ºC. The temperature drops by twice as much from Wednesday
Finger [1]

Answer: -8^{\circ}C

Step-by-step explanation:

Given

The temperature on Tuesday is 1^{\circ}C

Temperature drops to -2^{\circ}C on Wednesday

Net drop in temperature 1-(-2)=3^{\circ}C

The temperature drops twice from Wednesday to Thursday

\Delta T=2\times 3=6^{\circ}C

Temperature on Thursday

\Rightarrow -2-6=-8^{\circ}C

3 0
2 years ago
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