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Elan Coil [88]
3 years ago
15

The following sequence represents an arithmetic sequence. 1, 2, 3, 4, 1, 2, 3, 4, . . . true false

Mathematics
2 answers:
erma4kov [3.2K]3 years ago
8 0
False as the difference between two consecutive numbers is not a constant
Ainat [17]3 years ago
8 0

Answer:

False

Step-by-step explanation:

Given sequence,

1, 2, 3, 4, 1, 2, 3, 4, .........

Since, a sequence is called Arithmetic sequence if the difference in consecutive terms is constant,

∵ 2 - 1 = 3 - 2 = 4 - 3 = 1 - 4 ≠ 2 - 1 = 3 - 2 = 4 - 3 .......

Hence, the difference in consecutive terms is not constant,

Therefore, given sequence is not an Arithmetic sequence.

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What is the range of the function on the graph?
SIZIF [17.4K]

Answer: All real numbers less than or equal to 3.

Explanation: I took a test with this question and got it right.

3 0
3 years ago
What is 3:7 put in to lowest forms?
Artemon [7]
3/7 is in simplest form, because they are both prime numbers and have no factors.
4 0
3 years ago
Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.
Lelechka [254]

Answer:

1) \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4} = \frac{1}{32}

3) (-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}

4) \frac{2}{2^{-4}} = 32

Step-by-step explanation:

1) \frac{(-2)^{-5}}{(-2)^{-10}}

Solving using exponent rule: a^{-m}=\frac{1}{a^m}

\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32

So, \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4}

Using the exponent rule: a^m.a^n=a^{m+n}

We have:

2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}

We also know that: a^{-m}=\frac{1}{a^m}

Using this rule:

2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}

So, 2^{-1}.2^{-4} = \frac{1}{32}

3) (-\frac{1}{2} )^3.(-\frac{1}{2} )^2

Solving:

(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}

So, (-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}

4) \frac{2}{2^{-4}}

We know that: a^{-m}=\frac{1}{a^m}

\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32

So, \frac{2}{2^{-4}} = 32

3 0
3 years ago
Find x<br><br><br><br><br> Thanks for the help in advance!
Marina86 [1]

I'm not sure if this is the easiest way of doing this, but it surely work.

Let the base of the triangle be AB, and let CH be the height. Just for reference, we have

AH=2,\quad HB=6,\quad AC=x

Moreover, let CH=y and BC=z

Now, AHC, CHB and ABC are all right triangles. If we write the pythagorean theorem for each of them, we have the following system

\begin{cases}4+y^2=x^2\\36+y^2=z^2\\x^2+z^2=64\end{cases}

If we solve the first two equations for y squared, we have

y^2=x^2-4\\y^2=z^2-36

And we can deduce

z^2 = x^2+32

So that the third equation becomes

x^2+x^2+32=64 \iff 2x^2 = 32 \iff x^2=16 \iff x=4

(we can't accept the negative root because negative lengths make no sense)

7 0
2 years ago
I rent a gym for $150 for 30 students another time I rent the gym for $270 for 70 students I need to find a fixed rate
kirza4 [7]

Answer:

Fixed rate is $60.

Step-by-step explanation:

Let us consider per student charge be 'x'.

Let us consider fixed rate be 'b'

Given:

I rent a gym for $150 for 30 students.

So we can say that;

Total amount is equal to sum of number of students multiplied by per student charge and fixed rate.

framing in equation form we get;

30x+b=150 \ \ \ \ equation \ 1

Also Given:

another time I rent the gym for $270 for 70 students.

So we can say that;

Total amount is equal to sum of number of students multiplied by per student charge and fixed rate.

framing in equation form we get;

70x+b=270 \ \ \ \ equation \ 2

Now we will subtract equation 1 from equation 2 we get;

70x+b-(30x+b)=270-150\\\\70x+b-30x-b=120\\\\40x=120

Dividing both side by 40 we get;

\frac{40x}{40}=\frac{120}{40}\\\\x=\$3

Now we will substitute the value of x in equation 1 we get;

30x+b=150\\\\30\times3+b=150\\\\90+b=150\\\\b=150-90 =\$60

So we can say that the equation can be written as;

y =3x+60

Hence we can say that fixed rate is $60 and per student charge is $3.

3 0
3 years ago
Read 2 more answers
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