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3 years ago

The next number in the sequence -5, -2, 4, 13 is 25. Is this conjecture true or false?

Mathematics

1 answer:

steposvetlana [31]3 years ago

7 0

Answer:

False

Step-by-step explanation:

The sequence seems to be increasing by 3+3(-5+3=-2, -2+3+3=4, so on). By 13, the sequence is increasing by 11, making the next number 24.

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Drive 1,800 mi in 30 h

ella [17]

1800 mi / 30 hr = 60 mi/hr

hope that helped

6 0

3 years ago

Isolate the absolute value expression in the following equations to determine if they can be solved if so find and graph the sol

aev [14]

Answer:

x=-4

Step-by-step explanation:

2(|x+4|+3)=6

take out absolute value

2(x+4+3)=6

remove parrenthesis

2x+8+6=6

Simplify

2x+14=6

-14 -14

2x=-8

x=-4

Hope this helps!!!!!

4 0

3 years ago

Find the value of given expression <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B213%20%5Ctimes%20213%7D%20" id="TexForm

Hatshy [7]

Answer:

213 Answer ..............

7 0

3 years ago

Read 2 more answers

A researcher is studying the monthly gross incomes of drivers for a ride sharing company. (Gross incomes represent the amount pa

Reptile [31]

Answer:

The histogram of the sample incomes will follow the normal curve.

Step-by-step explanation:

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

In this case the researches wants to determine the monthly gross incomes of drivers for a ride sharing company.

He selects a sample of n = 200 drivers and ask them their monthly salary.

As the sample selected is quite large, i.e. n = 200 > 30, the central limit theorem can be applied to approximate the sampling distribution of sample mean by the Normal distribution.

Thus, the histogram of the sample incomes will follow the normal curve.

8 0

3 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.

7 0

3 years ago