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3 years ago

Write an expression with the five terms containing the coefficients 12 15 18 and 21

1 answer:

5 0

Let the variable used in the expression be "x".

Since the desired expression should contain 5 terms, therefore the powers of "x" will be as follows:
x^0 , x^1 , x^2, x^3 and x^4 where x^0 = 1 and x^1 = x

Now we are given only 4 coefficients to use, so we will have to assume a fifth assume.
Assume the fifth coefficient to be 7

It is a convention to write the powers in an expression in descending order, so the expression will be as follows (without coefficients):
x^4 + x^3 + x^2 + x + constant

Now add the coefficients in any desired order since there are no conditions, the final expression can be written as follows:
12 x^4 +15 x^3 +18 x^2 + 21 x + 7

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Plz help with this problem pic down below

kherson [118]

Answer:

yes is is

Step-by-step explanation:

if you connect the lines it forms a straight line

3 0

3 years ago

The average mass of an ant is approximately 3 × 10 − 3 grams. The average mass of a giraffe is approximately 2 × 10 3 kilograms.

Vika [28.1K]

Given that average mass of an ant $= 3 \times 10^{-3}$ grams.

Given that average mass of a giraffe $= 2 \times 10^{3}$ Kilograms.

Now we have to find about how many times more mass does a giraffe have than an ant. Before carring out any comparision, we must make both units equal.

Like convert kilogram into gram or gram into kilogram.

I'm going to convert kilogram into gram using formula

1 Kg = 1000 g

So the average mass of a giraffe $= 2 \times 10^{3} *10^3$ grams.

Now we just need to divide mass of giraffe by mass of ant to find the answer.

$\frac{2 \times 10^{3} *10^3 }{3 \times 10^{-3} }$

$= \frac{2 \times 10^{3} *10^3 *10^{3}}{3 }$

$= \frac{2 \times 10^{3+3+3}}{3 }$

$= \frac{2 \times 10^9}{3 }$

$= \frac{2}{3 } \times 10^9$

=666666666.667

Hence final answer is $\frac{2}{3 } \times 10^9$ which is approx 666666666.667.

3 0

3 years ago

At a fast food restaurant, 10 cheeseburgers w/onion and 5 orders of french fries contain 6900 calories. 5 cheeseburgers w/oni

Leona [35]

Write the equations,

10 c + 5 f = 6900

5c + 3f = 3660

Double,

10c + 6f = 7320

Subtract,

f = 7320 - 6900 = 420

10 c = 6900 - 5f = 6900 - 2100 = 4800

c = 480

Answer: Cheeseburger 480 calories, fries 420 calories

Check:

10 (480) + 5 (420) = 4800 + 2100 = 6900 good

5(480) + 3(420) = 2400 + 1260 = 3660 good

5 0

3 years ago

I need help.. i really want to go sleep.. thank you so much...

Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Now find $\sum\limits_{i=3}^{11}\frac{1}{i}$

Expanding the summation we get

$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Comparing the two series we get,

$\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$ so the given equality is true.

2) Given $\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}$

Verify the above equality is true or false

Now find $\sum\limits_{k=0}^4\frac{3k+3}{k+6}$

Expanding the summation we get

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}$

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}$

now find $\sum\limits_{i=1}^3\frac{3i}{i+5}$

Expanding the summation we get

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}$

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}$

Comparing the series we get that the given equality is false.

ie, $\sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}$

6 0

4 years ago

There are 12 face cards in a deck of standard playing cards and 20 even numbered cards (2; 4; 6; 8; 10 in each suit). what is th

FromTheMoon [43]

The probability of drawing a face card is 12/52
the probability of drawing an even number is 20/52
face cards and even numbered cards do not overwrap, so no need to subtract anything
Add up: 12/52+20/52=32/52=0.6154

5 0

3 years ago