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antiseptic1488 [7]
3 years ago
10

Given the functions f(x)= 1/x−4 +3 and g(x)=1/x+1 +6 .

Mathematics
2 answers:
melomori [17]3 years ago
7 0

ANSWER

The graph shifts 5 units left and 3 units up.

EXPLANATION

The given functions are:

f(x) =  \frac{1}{x - 4}  + 3

and

g(x) =  \frac{1}{x + 1}  + 6

We want to transform f(x) so that, it coincide with the g(x).

The required transformation is

f(x+5)+3

Let us see why this works.

f(x+5)+3 = \frac{1}{x + 5- 4}  + 3 + 3

f(x+5)+3 = \frac{1}{x + 1}  + 6 = g(x)

The transformation f(x+5)+3 will shift f(x) 5 units left and 3 units up.

The third choice is correct.

babunello [35]3 years ago
3 0

<u>Answer with step-by-step explanation:</u>

We are given a function f(x) =  \frac{1}{x - 4}  + 3 which is to be transformed to another function g(x) =  \frac{1}{x + 1}  + 6.

We are to determine whether which statement describes the transformation of the graph of function f onto the graph of function g.

f(x+5)+3 = \frac{1}{x + 5- 4}  + 3 + 3

f(x+5)+3 = \frac{1}{x + 1}  + 6

f(x)=g(x)

Therefore, the correct answer option is: The graph shifts 5 units right and 3 units up.

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If the quotient of 3/8 and 2 is subtracted from the product of 1 3/4 and 8/12 what is the difference please help
aleksandr82 [10.1K]

Answer:

47 / 48

Step-by-step explanation:

Quotient of 3/8 and 2

3/8 ÷ 2

3/8 * 1/2

= 3 /16

1 3/4 * 8/12

7/4 * 8/12

= 56/48

56/48 - 3/16

L. C. M of 48 and 16 = 48

(56 - 9) / 48

= 47 / 48

5 0
2 years ago
Triangle CDE is an isosceles triangle with angle d congruent to angle E if CD= 4x+9, DE= 7x-5, CE= 16x-27, find x and the measur
Montano1993 [528]
ANSWER

x = 3
CD= 21 \: units
DE= 16 \: units

CE= 21 \: units



EXPLANATION

It was given that ∆CDE is an isosceles triangle.

The length of the sides are given in terms of x as follows.

CD=4x+9


DE=7x-5


CE=16x-27



It was also given that, angle D is congruent to angle E.


This implies that, side CD and CE are equal.


Thus,

|CD|  =  |CE|
In terms of x, we have;

4x + 9 = 16x - 27


We group like terms to get,

16x - 4x = 9 + 27

12x = 36

Divide both sides by 12 to get,

x =  \frac{36}{12}

\therefore \: x = 3


The length of the sides are;

CD=4(3)+9 = 21 \: units

DE=7(3)-5 = 16 \: units


CE=16(3)-27 = 21 \: units
3 0
3 years ago
The first 6 nonzero multiples of 5
Vikentia [17]
<span>5: 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75.... </span>
5 0
3 years ago
PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS
Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\

With respect to reference angle A, we have:

  • opposite side = a = 4
  • adjacent side = b = \sqrt{33}
  • hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\

Then cosine which is adjacent over hypotenuse

\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\

Tangent is the ratio of opposite over adjacent

\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

  • cosecant, abbreviated as csc, is the reciprocal of sine
  • secant, abbreviated as sec, is the reciprocal of cosine
  • cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{  ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

------------------------------------------------------

Summary:

The missing side is b = \sqrt{33}

The 6 trig functions have these results

\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0
1 year ago
2+2^2 divided by 4+3
denis-greek [22]

Answer:

4.5

Step-by-step explanation:

7 0
3 years ago
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