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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)

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