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lukranit [14]
2 years ago
15

PLEASE ANSWER ASAPP I HAVE A PICTURE ATTACHED

Mathematics
2 answers:
Alik [6]2 years ago
3 0
Its c I've done this before
victus00 [196]2 years ago
3 0

Answer:

C

Step-by-step explanation:

just do it.

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There are 600 cars in the parking lot. The ratio of red cars to white cars is 3:5. How many red cars and how many white cars are
kicyunya [14]

Answer:

360 red cars

Step-by-step explanation:

your welcome (yw)

8 0
3 years ago
Find the unit rate. 256 miles per 8 gallons
denpristay [2]
<span>the answer is 32 miles per 1 gallon.</span>
3 0
2 years ago
Read 2 more answers
Need help on all these questions!!
Yuki888 [10]

Okay, so YZ = 3 cm.  You have XM correct. And YM = 0.5.

Now, you have the midpoint M at the correct spot.

Use Pythagorean's theorem o find the length of AB.  a² + b² = c²   a=6,  b=8.

6² = 36   8² = 64    36 + 64 = 100      AB = 10!

If AB = 10   then AM = 5    MB also = 5

If B is the midpoint of AC,  C would be 12 rows down from A, and 16 columns to the right. The last spot where the line intersects.

There are your answers!

5 0
2 years ago
What is the slope of (-2,7); (-5,3)
Vinvika [58]

Answer:

1 1/3

Step-by-step explanation:

m = y2-y1 over x2-x1

3 0
2 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
2 years ago
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