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gregori [183]
3 years ago
6

If you have 5 eyes what eoul you do?​

Mathematics
1 answer:
postnew [5]3 years ago
5 0

Answer:

im in to bullying, because i have 5 eyes i know it's is so problem to 5 eye but don't be affairs to people how hates you,don't be shy be positive with yourself and the people how hates you will be apologized to you.

Step-by-step explanation:

i hope its help you

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Thirty- six shirts were on sale for 2/3 of the regular price of the sale price of each shirt was 30 what was the regular price?
Vika [28.1K]
First change 2/3 into a decimal: .67. Then divide 30 by .67: 44.77. Round it: 44.77 = 45. Each shirt was $45 :)
5 0
3 years ago
(I need help. I’m having trouble understanding Plz And I'm giving brainly thanks
lesya [120]

The answer will be 12413. Please mark me Brainliest

7 0
3 years ago
How can I reflect this horizontally without a value for h?
Alex_Xolod [135]

Answer:

Another transformation that can be applied to a function is a reflection over the x– or y-axis. A vertical reflection reflects a graph vertically across the x-axis, while a horizontal reflection reflects a graph horizontally across the y-axis. The reflections are shown in Figure 9.

Graph of the vertical and horizontal reflection of a function.

Figure 9. Vertical and horizontal reflections of a function.

Notice that the vertical reflection produces a new graph that is a mirror image of the base or original graph about the x-axis. The horizontal reflection produces a new graph that is a mirror image of the base or original graph about the y-axis.

A GENERAL NOTE: REFLECTIONS

Given a function \displaystyle f\left(x\right)f(x), a new function \displaystyle g\left(x\right)=-f\left(x\right)g(x)=−f(x) is a vertical reflection of the function \displaystyle f\left(x\right)f(x), sometimes called a reflection about (or over, or through) the x-axis.

Given a function \displaystyle f\left(x\right)f(x), a new function \displaystyle g\left(x\right)=f\left(-x\right)g(x)=f(−x) is a horizontal reflection of the function \displaystyle f\left(x\right)f(x), sometimes called a reflection about the y-axis.

HOW TO: GIVEN A FUNCTION, REFLECT THE GRAPH BOTH VERTICALLY AND HORIZONTALLY.

Multiply all outputs by –1 for a vertical reflection. The new graph is a reflection of the original graph about the x-axis.

Multiply all inputs by –1 for a horizontal reflection. The new graph is a reflection of the original graph about the y-axis.

EXAMPLE 7: REFLECTING A GRAPH HORIZONTALLY AND VERTICALLY

Reflect the graph of \displaystyle s\left(t\right)=\sqrt{t}s(t)=√

t

(a) vertically and (b) horizontally.

SOLUTION

a. Reflecting the graph vertically means that each output value will be reflected over the horizontal t-axis as shown in Figure 10.

Graph of the vertical reflection of the square root function.

Figure 10. Vertical reflection of the square root function

Because each output value is the opposite of the original output value, we can write

\displaystyle V\left(t\right)=-s\left(t\right)\text{ or }V\left(t\right)=-\sqrt{t}V(t)=−s(t) or V(t)=−√

t

Notice that this is an outside change, or vertical shift, that affects the output \displaystyle s\left(t\right)s(t) values, so the negative sign belongs outside of the function.

b.

Reflecting horizontally means that each input value will be reflected over the vertical axis as shown in Figure 11.

Graph of the horizontal reflection of the square root function.

Figure 11. Horizontal reflection of the square root function

Because each input value is the opposite of the original input value, we can write

\displaystyle H\left(t\right)=s\left(-t\right)\text{ or }H\left(t\right)=\sqrt{-t}H(t)=s(−t) or H(t)=√

−t

Notice that this is an inside change or horizontal change that affects the input values, so the negative sign is on the inside of the function.

Note that these transformations can affect the domain and range of the functions. While the original square root function has domain \displaystyle \left[0,\infty \right)[0,∞) and range \displaystyle \left[0,\infty \right)[0,∞), the vertical reflection gives the \displaystyle V\left(t\right)V(t) function the range \displaystyle \left(-\infty ,0\right](−∞,0] and the horizontal reflection gives the \displaystyle H\left(t\right)H(t) function the domain \displaystyle \left(-\infty ,0\right](−∞,0].

TRY IT 2

Reflect the graph of \displaystyle f\left(x\right)=|x - 1|f(x)=∣x−1∣ (a) vertically and (b) horizontally.

Solution

EXAMPLE 8: REFLECTING A TABULAR FUNCTION HORIZONTALLY AND VERTICALLY

A function \displaystyle f\left(x\right)f(x) is given. Create a table for the functions below.

\displaystyle g\left(x\right)=-f\left(x\right)g(x)=−f(x)

\displaystyle h\left(x\right)=f\left(-x\right)h(x)=f(−x)

\displaystyle xx 2 4 6 8

\displaystyle f\left(x\right)f(x) 1 3 7 11

SOLUTION

For \displaystyle g\left(x\right)g(x), the negative sign outside the function indicates a vertical reflection, so the x-values stay the same and each output value will be the opposite of the original output value.

\displaystyle xx 2 4 6 8

\displaystyle g\left(x\right)g(x) –1 –3 –7 –11

For \displaystyle h\left(x\right)h(x), the negative sign inside the function indicates a horizontal reflection, so each input value will be the opposite of the original input value and the \displaystyle h\left(x\right)h(x) values stay the same as the \displaystyle f\left(x\right)f(x) values.

\displaystyle xx −2 −4 −6 −8

\displaystyle h\left(x\right)h(x) 1 3 7 11

TRY IT 3

\displaystyle xx −2 0 2 4

\displaystyle f\left(x\right)f(x) 5 10 15 20

Using the function \displaystyle f\left(x\right)f(x) given in the table above, create a table for the functions below.

a. \displaystyle g\left(x\right)=-f\left(x\right)g(x)=−f(x)

b. \displaystyle h\left(x\right)=f\left(-x\right)h(x)=f(−x)

3 0
2 years ago
How do you do 4x+27=3x??
KATRIN_1 [288]
4x + 27 = 3x

First, take away '4x' from both sides.
27 = 3x - 4x
Second, subtract '3x - 4x' to get '-x'.
27 = -x
Third, multiply both sides by '-1'.
-27 = x
Fourth, switch your sides.
x = -27

Answer: x = -27

7 0
3 years ago
Read 2 more answers
Show that y=2e^(−x) cos(x)−e^(−x) sin(x) is a solution to y''+2y'+2y=0.
Furkat [3]

Answer:

y=2e^(−x)cosx−e^(−x)sinx

Satisfies the equation

Step-by-step explanation:

Answer:

y=2e^(−x)cosx−e^(−x)sinx

y = e^(-x)[2cosx - sinx]

Find y' and y" using product law

y' = -e^(-x)[2cosx - sinx] + e^(-x)[-2sinx - cosx]

y' = -e^(-x)[2cosx - sinx + 2sinx + cosx]

y' = -e^(-x)[3cosx + sinx]

y" = e^(-x)[3cosx + sinx] - e^(-x)[-3sinx + cosx]

y" = e^(-x)[3cosx - cosx + sinx + 3sinx]

y" = e^(-x)[2cosx + 4sinx]

y" + 2y' + 2y

e^(-x)[2cosx + 4sinx] - 2e^(-x)[3cosx + sinx] +2e^(-x)[2cosx - sinx]

e^(-x)[4sinx - 2sinx - 2sinx + 2cosx - 6 cosx + 4cosx]

= e^(-x) × 0

= 0

7 0
4 years ago
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