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Digiron [165]
3 years ago
12

PLZ HELP!!! if p = -1,-1 find the image of p under the following rotation 270 counterclockwise

Mathematics
1 answer:
seraphim [82]3 years ago
3 0
The image of p is (-1, 1)
Because the rule for 270 counterclockwise is (y,-x)
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A point R (-2,5) is mapped to point R',as described by a translation of 6 units to the rights and 7 units down.Mark R' on the co
laila [671]

Answer:

(4, -2)

Step-by-step explanation:

Given a point R (-2,5), if the point R' is described by a translation of 6 units to the rights and 7 units down, then the coordinate of R' will be;

6 units to the rights is towards the positive x axis

7 units down is towards the negative y axis

R' = (-2+6, 5-7)

R' = (4, -2)

Hence the coordinate point of R' on the plane is (4, -2)

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3 years ago
Help me on this please
KonstantinChe [14]

Answer:

14.387

Step-by-step explanation:

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The function f is linear. If f(2) = 2 and f(-6)=6 find f(6)
lina2011 [118]
The\ slope-intercept\ form:y=mx+b\\\\f(2)=2\to for\ x=2\to y=2\\f(-6)=6\to for x=-6\to y=6\\\\put\ in\ equation\ of\ the\ line:\\\\  -\left\{\begin{array}{ccc}2=2m+b\\6=-6m+b\end{array}\right\ \ \ |subtract\ sides\ of\ the\ equations\\----------\\.\ \ \ -4=8m\ \ \ \ \ |divide\ both\ sides\ by\ 8\\.\ \ \ \ \boxed{m=-\frac{1}{2}}\\\\put\ m=-\frac{1}{2}\ to\ equation\ 2=2m+b:\\\\2=2\cdot(-\frac{1}{2})+b\\2=-1+b\ \ \ \ |add\ 1\ to\ both\ sides\\\boxed{b=3}\\\\\boxed{\boxed{y=-\frac{1}{2}x+3}}

therefore:f(x)=-\frac{1}{2}x+3\Rightarrow f(6)=-\frac{1}{2}\cdot6+3=-3+3=0\\\\Answer:\boxed{f(6)=0}
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3 years ago
Find a number that is divisible by 2,3,5,6,9,and 10
FinnZ [79.3K]
<em>90</em> and all of its multiples are.
4 0
3 years ago
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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
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