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3 years ago

Consider the expression? Please help me !

Mathematics

2 answers:

Arada [10]3 years ago

8 0

Answer:

sqrt(-8x+5)

Step-by-step explanation:

(5-8x)

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

sqrt(-8x+5)

We need to rationalize the denominator

(5-8x) sqrt(-8x+5)

----------------- * ---------------

sqrt(-8x-5) sqrt(-8x+5)

(5-8x) sqrt(-8x+5)

----------------- * ---------------

(-8x+5)

The first term cancels

sqrt(-8x+5)

dmitriy555 [2]3 years ago

4 0

Answer:

Second option

Step-by-step explanation:

The radical represents the square root, i.e power ½

(5 - 8x) ÷ (5 - 8x)^½

(5 - 8x)^(1-½)

(5 - 8x)^½

(-8x + 5)^½ (just a rearrangement)

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Find f(-11/5) if f(n) = 5n + 6

V125BC [204]

Answer:

f(-11/5) = -5

Step-by-step explanation:

f(n) = 5n + 6

Let x = -11/5

f(-11/5) = 5*-11/5 +6

= -11 +6

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3 years ago

If h(x) = 5 x and k(x)=1/x, which expression is equivalent to (koh)(x)?

Lyrx [107]

May be there is an operator missing in the first function, h(x). I will solve this in two ways, 1) as if the h(x) = 5x and 2) as if h(x) = 5 + x

1) If h(x) = 5x and k(x) = 1/x

Then (k o h) (x) = k ( h(x) ) = k(5x) = 1/(5x)

2) If h(x) = 5 + x and k (x) = 1/x

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3 years ago

Read 2 more answers

If Y varies directly with X solve the following :

tresset_1 [31]

Answer:

B) When Y = 90

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= -270

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D) When Y = 12.6

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Hope this helps!

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3 years ago

Exploring the graph of y equals cosine Theta

irga5000 [103]

Answer: see below

Step-by-step explanation:

The coordinates on the Unit Circle are (cos, sin). Since we are focused on cosine, we only need to focus on the left side of the coordinate. The cosine value (left side) will be the y-value of the function y = cos x

Use the quadrangles (angles on the axes) to represent the x-values of the function y = cos x.

Quadrangles are: 0°, 90°, 180°, 270°, 360° (360° = 0°)

Together, the coordinates will be as follow:

$\boxed{\begin{array}{c||c|c||c}\underline{(cos,sin)}&\underline{x=angle}&\underline{y=cosx}&\underline{\quad (x,y)\quad}\\(1,0)&0^o&1&(0^o,1)\\(0,1)&90^o&0&(90^o,0)\\(-1,0)&180^o&-1&(180^o,-1)\\(0,-1)&270^o&0&(270^o,0)\\(1,0)&360^o&1&(360^o,1)\end{array}}$

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3 years ago