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3241004551 [841]
3 years ago
12

What is 31/4 + -11/2 + 21/4

Mathematics
2 answers:
Basile [38]3 years ago
6 0
31/4 + -11/2 + 21/4 = 7 1/2
Hope this helps!
Andrei [34K]3 years ago
6 0
Hope this helped, and GOOD LUCK! :D

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Write a linear equation that contains the points shown in the table.
Afina-wow [57]

Answer:

y=7x+1

Step-by-step explanation:

From the table, the difference in y is 7 and the difference in x is 1.

Let the linear equation be

y = mx + b

Where m is the constant difference in y divided by the constant difference in x.

This means m=7/1=7

Our equation now becomes:

y=7x+b

To find b, we substitute any ordered pair from the table.

From the to table, when x=1, y=8.

This implies that,

8=7(1)+b

b=8-7=1

Therefore the equation is y=7x+1

4 0
3 years ago
How many complex zeros does the polynomial function have?<br> f(x)=−3x^6−2x^4+5x+6
REY [17]

one way would be to factor

I can't factor it so we will have to use Descartes' Rule of Signs which is helpful for finding how many real roots you have


it goes like this:

for a polynomial with real coefients, consider f(x)=-3x^6-2x^4+5x+6.

after arranging the terms in decending order in terms of degree, count how many times the signs of the coeffients change direction and minus 2 from that number until you get to 1 or 0. that will be the number of even roots the function can have

We have (-, -, +, +). the signs changed 1 times, so it has 1 real positive root


to get the negative roots, we evaluate f(-x) and see how many times the root changes

f(-x)=-3x^6-2x^4-5x+6

signs are (-, -, -, +). there was 1 change in sign

so the function has 1 real negative root



a total of 2 real roots


a function of degree n can have at most, n roots


our function is degree 6 so it has 6 roots

if 2 are real, then the others must be complex

6-2=4 so there are 4 complex roots


you can also show that there are only 2 real roots by using a graphing utility to see that there are only 2 real roots

7 0
3 years ago
A person wishes to mix coffee worth nine dollars per pound with coffee worth three dollars per pound to get 240 pounds of a mixt
Anastasy [175]
\bf \begin{array}{lccclll}&#10;&amount&price&priced\ amount\\&#10;&-----&-----&--------\\&#10;\textit{9 bucks/lb}&x&9&9x\\&#10;\textit{3 bucks/lb}&y&3&3y\\&#10;-----&-----&-----&-------\\&#10;mixture&240&5&1200&#10;\end{array}&#10;\\\\\\&#10;\begin{cases}&#10;x+y=240\implies \boxed{y}=240-x\\&#10;9x+3y=1200\\&#10;----------\\&#10;9x+3\left( \boxed{240-x} \right)=1200&#10;\end{cases}

solve for "x".

what about "y"? well, y = 240 - x
4 0
3 years ago
Simplify the expression where possible. (r 3) -2
Julli [10]
What you do to on side of an equation (=) you must do to the other to keep both sides equal.

r3-2+2=+2  (add 2 to both sides of the equation to simplify the left side)
which becomes
r3=2
r3/3 = 2/3 (divide both sides by 3) Note 3/3 =1 and 1 r is the same as r
which becomes
r=2/3 .
5 0
3 years ago
A password contains six digits, such as 174668. How many different passwords can be formed
STALIN [3.7K]

Answer: The number of different passwords can be formed = 1,000,000

Step-by-step explanation:

In general total digits = 10  [ from 0 to 9]

If a password contains 6 digits, then by fundamental principal of counting, the total number of different passwords =10\times10 \times10\times10\times10\times10=10^6=1000000

Hence, the number of different passwords can be formed = 1,000,000

5 0
3 years ago
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