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2 years ago

6

What is (6,2) rotated at 270 .

1 answer:

erma4kov [3.2K]2 years ago

8 0

Answer:

55

Step-by-step explanation:

bc

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Helpppp please people

Arte-miy333 [17]

Answer:

3

Step-by-step explanation:

p-(9-(m+q)) =

5-(9-(4+3)) =

5-(9-(7)) =

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3

3 0

2 years ago

Read 2 more answers

Given the function, check all transformations that occurred from the graph of the function y=x^2

Hatshy [7]

Answer: left 7 units, up 4 units, and vertical compression

Step-by-step explanation:

Given

The function is $y=x^2$

First shift it to the left by 7 units, so the function becomes

$y=(x+7)^2$

Now compress it to make it one-third

$y=\dfrac{1}{3}(x+7)^2$

Now, sift the graph upward by 4 units

$y=\dfrac{1}{3}(x+7)^2+4$

Thus , the conditions are left 7 units, up 4 units, and vertical compression.

4 0

2 years ago

200 centimeters to 5 meters write ratio in simplest form

ss7ja [257]

Answer:

5.2

have a nice day

5 0

2 years ago

Use the intermediate value theorem to find the value of c such that f(c) = M. f(x) = x^2 - x + 1 text( on ) [1,8]; M = 21 c =

DanielleElmas [232]

Answer:

$c = 5$

Step-by-step explanation:

Given

$f(c) = M$

$f(x) = x^2 - x + 1$

$Interval: [1,8]$

$M = 21$

Required

Find c using Intermediate Value theorem

First, check if the value of M is within the given range:

$f(x) = x^2 - x + 1$

$f(1) = 1^2 - 1 + 1$

$f(1) = 1$

$f(x) = 8^2 - 8 + 1$

$f(x) = 57$

$1 \le M \le 57$

$1 \le 21 \le 57$

M is within range.

Solving further:

We have:

$f(c) = f(x) = M$

$f(x) = 21$

Substitute 21 for f(x) in $f(x) = x^2 - x + 1$

$21 = x^2 - x + 1$

Express as quadratic function

$x^2 - x + 1 - 21 = 0$

$x^2 - x - 20 = 0$

Expand

$x^2 + 4x - 5x - 20$

$x(x+4)-5(x+4)=0$

$(x - 5)(x+4) = 0$

$x - 5 = 0$ or $x + 4= 0$

$x = 5$ or $x = -4$

The value of $x = -4$ is outside the $Interval: [1,8]$

So:

$x = 5$

$f(c) = f(x) = M$

$f(c) = f(5) = 21$

By comparison:

$c = 5$

4 0

2 years ago

Solve this recurrence relation<br><br>an= -an-2<br>a1=1<br>a2=2

Lesechka [4]

Answer:

Give Full Question ...I think this question is incomplete

3 0

2 years ago