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Illusion [34]
4 years ago
7

steve has these score on the exams for this semester 89 92 and 87. How much will have to make on the fourth exam to have a final

average of an A( 92% or better)

Mathematics
1 answer:
Zinaida [17]4 years ago
6 0
THIS IS THE CORRECT ANSWER I GOT WHEN SOLVING ON MY IPHONE

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If you rearrange the equation so w is the independent variable, then what is u?<br><br> -2u+6w=9
Serjik [45]

Answer:

u  = - ( 9-6w)/2

Step-by-step explanation:

to  evaluate for u in the expression -2u+6w=9 is simply to look for a way such that we would express u in terms of w and other variables.

solution

-2u+6w=9

-2u = 9 - 6w

divide both sides by the coefficient of u which is -2

-2u/u = 9 - 6w/-2

u  = - ( 9-6w)/2

therefore the value of u when rearranged in the equation -2u+6w=9 is evaluated to be equals to

u  = - ( 9-6w)/2

7 0
3 years ago
Determine the vertical asymptote for the rational function f(x) = x - 4 over 2x - 3
k0ka [10]

Answer:

x = \frac{3}{2}

Step-by-step explanation:

Given

f(x) = \frac{x-4}{2x-3}

The denominator cannot be zero as this would make f(x) undefined.

Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non zero for this value then it is a vertical asymptote.

2x - 3 = 0 ⇒ 2x = 3 ⇒ x = \frac{3}{2}

Thus x = \frac{3}{2} is the vertical asymptote

5 0
3 years ago
Which pair of terms are like terms?
sdas [7]

Answer:

b: -2y and 4y

Step-by-step explanation:

The answer is option B

-2y and 4y are like terms because they have the same algebraic alphabet at the back of their coefficients.

6 0
3 years ago
Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______
Ludmilka [50]

Answer:

(a)\ \sec^2(\theta) = 82

(b)\ \cot(\theta) = \frac{1}{9}

(c)\ \cot(\frac{\pi}{2} - \theta) = 9

(d)\ \csc^2(\theta) = \frac{82}{81}

Step-by-step explanation:

Given

\tan(\theta) = 9

Required

Solve (a) to (d)

Using tan formula, we have:

\tan(\theta) = \frac{Opposite}{Adjacent}

This gives:

\frac{Opposite}{Adjacent} = 9

Rewrite as:

\frac{Opposite}{Adjacent} = \frac{9}{1}

Using a unit ratio;

Opposite = 9; Adjacent = 1

Using Pythagoras theorem, we have:

Hypotenuse^2 = Opposite^2 + Adjacent^2

Hypotenuse^2 = 9^2 + 1^2

Hypotenuse^2 = 81 + 1

Hypotenuse^2 = 82

Take square roots of both sides

Hypotenuse =\sqrt{82}

So, we have:

Opposite = 9; Adjacent = 1

Hypotenuse =\sqrt{82}

Solving (a):

\sec^2(\theta)

This is calculated as:

\sec^2(\theta) = (\sec(\theta))^2

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

Where:

\cos(\theta) = \frac{Adjacent}{Hypotenuse}

\cos(\theta) = \frac{1}{\sqrt{82}}

So:

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2

\sec^2(\theta) = (\sqrt{82})^2

\sec^2(\theta) = 82

Solving (b):

\cot(\theta)

This is calculated as:

\cot(\theta) = \frac{1}{\tan(\theta)}

Where:

\tan(\theta) = 9 ---- given

So:

\cot(\theta) = \frac{1}{\tan(\theta)}

\cot(\theta) = \frac{1}{9}

Solving (c):

\cot(\frac{\pi}{2} - \theta)

In trigonometry:

\cot(\frac{\pi}{2} - \theta) = \tan(\theta)

Hence:

\cot(\frac{\pi}{2} - \theta) = 9

Solving (d):

\csc^2(\theta)

This is calculated as:

\csc^2(\theta) = (\csc(\theta))^2

\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2

Where:

\sin(\theta) = \frac{Opposite}{Hypotenuse}

\sin(\theta) = \frac{9}{\sqrt{82}}

So:

\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2

\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2

\csc^2(\theta) = \frac{82}{81}

4 0
3 years ago
The size of the left upper chamber of the heart is one measure of cardiovascular health. When the upper left chamber is enlarged
Ganezh [65]

Answer:

a) P [ X < 24 mm ] = 0,3015          or     P [ X < 24 mm ] =  30,15 %

b) P [ X > 32 mm ]  = 0,1251          or        P [ X > 32 mm ]  = 12,51 %

c) P [  25 < X < 30 ] = 0,4964      or     P [  25 < X < 30 ] = 49,64 %

d) z(s) = 0,84

Step-by-step explanation:

Normal Distribution    N (  μ₀  ;   σ )   is  N ( 26,5 ; 4,8 )

a) P [ X < 24 mm ] = ( X - μ₀ ) / σ

P [ X < 24 mm ] = (24 - 26,5)/ 4,8 = - 0,5208 ≈ - 0,52

P [ X < 24 mm ] = - 0,52  

And from z-table we find area for z score

P [ X < 24 mm ] = 0,3015          or     P [ X < 24 mm ] =  30,15 %

b)P [ X > 32 mm ]  =  1  - P [ X < 32 mm ]

P [ X < 32 mm ]  = ( 32 - 26,5 ) / 4,8

P [ X < 32 mm ]  = 5,5/4,8  =  1,1458 ≈ 1,15

P [ X < 32 mm ]  = 1,15

And from z-table  we get

P [ X < 32 mm ]  = 0,8749

Then:

P [ X > 32 mm ]  =  1  -  0,8749

P [ X > 32 mm ]  = 0,1251          or        P [ X > 32 mm ]  = 12,51 %

c) P [  25 < X < 30 ] = P [ X < 30 ] - P [ X < 25 ]

P [ X < 30 ]  = 30 - 26,5 / 4,8   =  0,73

From  z-table    P [ X < 30 ]  =  0,7673

P [ X < 25 ] = 25 - 26,5 / 4,8  = - 0,3125  ≈  - 0,31

From z-table  P [ X < 25 ] =  0,2709

Then

P [  25 < X < 30 ] =  0,7673 -  0,2709

P [  25 < X < 30 ] = 0,4964      or     P [  25 < X < 30 ] = 49,64 %

d) If  20 %

z- score for 20% is from z-table

z(s) = 0,84

6 0
3 years ago
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