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AysviL [449]
3 years ago
6

- Save & Exit Certify

Mathematics
1 answer:
OverLord2011 [107]3 years ago
4 0

Answer:

Do you have a picture because i work better with pictures

Step-by-step explanation:

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M=1/6yd solve for y???
8090 [49]
6Myd = 1

y= 1/6Md is the answer
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3 years ago
What is y to the 5th power divided by y to the 8th power?
swat32
All you have to do is divided the other two numbers and than find out what y I and divide that than you will get it
5 0
3 years ago
Read 2 more answers
Which of the following relations represent a function? A. {(-1,-1),(0,0),(2,2),(5,5)}. B. None of these. C. {(0,3),(0,-3),(-3,0)
Cerrena [4.2K]

Answer:

  A

Step-by-step explanation:

In a list of ordered pairs, a relation that is a function will not have any repeated x-values.

A -- a function (x-values are {-1, 0, 2, 5} with no repeats)

B -- irrelevant

C -- not a function (x-values are {0, 0, -3, 3} with 0 being repeated)

D -- not a function (x-values are {-2, -1, -2, 2} with -2 being repeated)

3 0
3 years ago
2.6 yards to inches!? Explain it
motikmotik

Answer: 93.6 inches

there are 36 inches is a yard so 2.6 x 36 = 93.6

Step-by-step explanation:

4 0
3 years ago
Given the points C(-1,-5) and D(-7,11) find the coordinates of point E on CD such that the ratio of CE to ED is 5:3.
Anon25 [30]

Answer:

The coordinates of point E on CD are \left(-\frac{19}{4},5 \right).

Step-by-step explanation:

Let be C = (-1,-5), D = (-7,11) and \frac{CE}{ED} = \frac{5}{3}. The given ratio can be translated vectorially into this:

\overrightarrow {CE} = \frac{5}{3} \cdot \overrightarrow{ED}

And let consider that each point is a vector with respect to origin:

\vec C = (-1, -5) and \vec {D} = (-7,11)

Then,

\vec E -\vec C = \frac{5}{3}\cdot (\vec D -\vec E)

\vec E +\frac{5}{3}\cdot \vec E = \frac{5}{3}\cdot \vec D + \vec C

\frac{8}{3}\cdot \vec E = \frac{5}{3}\cdot \vec D + \vec C

\vec E = \frac{5}{8}\cdot \vec D + \frac{3}{8}\cdot \vec C

\vec E = \frac{5}{8}\cdot (-7,11)+\frac{3}{8}\cdot (-1,-5)

\vec E = \left(-\frac{35}{8},\frac{55}{8}  \right)+\left(-\frac{3}{8},-\frac{15}{8}  \right)

\vec E = \left(-\frac{35}{8}-\frac{3}{8},\frac{55}{8}-\frac{15}{8}    \right)

\vec{E} = \left(-\frac{19}{4}, 5\right)

The coordinates of point E on CD are \left(-\frac{19}{4},5 \right).

6 0
3 years ago
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