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Sav [38]
3 years ago
6

To shift the graph of an equation a certain number of units _____, you need to subtract that number from the functions equation

Mathematics
2 answers:
vredina [299]3 years ago
8 0

Answer: D, because A function can be shift vertically up or down by adding some positive or negative constant.

FromTheMoon [43]3 years ago
3 0

Answer:

The correct option is A) Down.

Step-by-step explanation:

A function can be shift vertically up or down by adding some positive or negative constant.

For example:

Consider the figure 1:

The function g(x)=f(x)+k shifted up by <em>k</em> unit. (where <em>k</em> is 1 unit)

Now consider the figure 2:

The function g(x)=f(x)-k shifted down by <em>k</em> unit. (where <em>k</em> is 1 unit)

Which means to shift the graph of an equation certain number of units up, we need to add that number from the functions equation.

Similarly, to shift the graph of an equation certain number of units down, we need to subtract that number from the functions equation.

Therefore, the correct option is A) Down.

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Which expressions are equivalent to 5^2/5^8?
ICE Princess25 [194]

Answer:

1/15625 I am pretty sure it is right but please double check.

Step-by-step explanation:

5^2=25

5^8=390625

7 0
3 years ago
Write an equation in slope-intercept form of the line that passes throught the point and given slope:
ludmilkaskok [199]
 y = mx + b
5 = 12(-3) + b
5 = -36 + b
b = 41

y = -3x + 41
6 0
3 years ago
Twelve percent of yesterday’s customers purchased premium-grade gasoline from GasCo. If 24 customers bought premium-grade gasoli
Oduvanchick [21]

Answer:

196

Step-by-step explanation:

100 divided by 12 is 8 r 4 24 x 8 = 192, 192 + 4 is 196

7 0
3 years ago
Unsure how to do this calculus, the book isn't explaining it well. Thanks
krok68 [10]

One way to capture the domain of integration is with the set

D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}

Then we can write the double integral as the iterated integral

\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx

Compute the integral with respect to y.

\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)

Compute the remaining integral.

\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}

We could also swap the order of integration variables by writing

D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}

and

\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy

and this would have led to the same result.

\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)

\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)

7 0
1 year ago
Omg plsss someone help i rly don't get this question!!! TYY
finlep [7]
Ok sorry but I see someone is online can you help
With a science question
4 0
3 years ago
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