For the following exercise, calculate the desired dose. Then calculate the amount to administer. Ordered: Pergolide mesylate 100 mcg PO tid On hand: Pergolide mesylate 0.05 mg tablets what is the Desired dose?​

Answer:

7/10 or 0.7

Step-by-step explanation:

a probability is always the ratio of possible cases over all cases.

"all cases" here is 100.

possible cases are all iPads not cracked in the inventory = 70 (because 30 are cracked, that leaves 100-30=70 not cracked).

so, the probability to select a non-cracked unit is

70/100 or simplified 7/10 (or 0.7)

Answer:

49,050cents

Step-by-step explanation:

Given the expression for calculating the marginal cost per card of producing x note cards expressed as

C'(x) = -0.05x + 77

On intergrating the marginal cost, we will get the total cost

C(x) =  -0.05x²/2 + 77x

Substitute x = 900 into the resulting expression

C(900) =  -0.05(900)²/2 + 77(900)

C(900) = -20,250+69,300

C(900) = 49,050

Hence the total costs in cents is 49,050cents

Answers:

Refer to the graph below.

==========================================================

Explanation:

If f(x) is an even function, then f(-x) = f(x) for all x in the domain.

What this means is that we have symmetry about the y axis. We can reflect that given curve over the y axis to generate the missing left side.

The graph shows that (1,1) is on the orange curve. It reflects over to (-1,1). This means 1 goes in the first box.

Use the rule [tex](x,y) \to (-x,y)[/tex] to apply a y axis reflection. We simply just change the sign of the x coordinate from positive to negative, while keeping the y coordinate the same.

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

We can also see that (3,1) is also on the orange curve. It reflects over to (-3, 1) using that rule mentioned earlier.

1 goes in the second box

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

The graph your teacher gave you shows that if we plugged in x = 5, then we get y = 3. In other words, the point (5,3) is on the orange graph.

It reflects over to (-5, 3) to show that x = -5 leads to the output y = 3

3 goes in the third box

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

Lastly, the point (6,3) reflects to (-6,3) when reflecting over the y axis.

3 goes in the fourth box.

See the graph below.

Answer:

2666.01%

Step-by-step explanation:

Intrest=(PA)*I*T

1377.53=51.67*I*1

I=26.6601=2666.01%

Answer:

9

Step-by-step explanation:

First multiple of 8 in that range is 8(2)=16.

The last multiple of 8 in that range is 8(10)=80.

So we just need to find how many numbers there are between 2 and 10. inclusive.

10-2+1=9

It's also not that long to write out and count.

1-2

2-3

3-4

4-5

5-6

6-7

7-8

8-9

9-10

9 numbers there are

Multiply those 9 numbers by i you will have all multiples of 8 btw 15 and 85.

8(2)=16

8(3)=24

8(4)=32

8(5)=40

8(6)=48

8(7)=56

8(8)=64

8(9)=72

8(10)=80

Answer:

m<GEF = 66°

Step-by-step explanation:

(72+60)/2

= 132/2

= 66

Answered by GAUTHMATH

Answer:

We have 60% salt and 85% salt solutions and we need 140 ounces of a solution that is 80% salt.

We set up 2 equations:

A) x + y = 140

B) .60x + .85y = .80 * 140

Multiply equation A) by -.60

A) -.60x -.60y = -84  then we add this to B)

B) .60x + .85y = 112

.25y = 28

y = 112 ounces of 85% salt

x = 28 ounces of 60% salt.

Double Check

112 * .85 = 95.2 AND 28 * .60 = 16.80

95.2 + 16.80 = 112

and 112/140 = 80 per cent!!

Source: http://www.1728.org/mixture.htm

Step-by-step explanation:

Answer:

21

Step-by-step explanation:

All three sides are equal

2x-7 = x+y-9 = y+5

Using the last two

x+y-9 = y+5

Subtract y from each side

x+y-9-y = y+5-y

x-9 = 5

Add 9 to each side

x -9+9 = 5+9

x=14

We know the side length is

2x-7

2(14) -7

28-7

21

The side length is 21

Answer:

106 5/6

Step-by-step explanation:

55 < 66 < 74 < 89 < 106

the fractions are irrelevant because the whole numbers are all different

The greatest of the numbers will be 106 5/6 since 106 is greater than the rest of the integers present in the mixed fractions.

The greatest value is the highest value in a given set of number.

Given the following numbers

89 3/4

66 1/2

55 2/3

74 1/4

106 5/6

The greatest of the numbers will be 106 5/6 since 106 is greater than the rest of the integers present in the mixed fractions.

Learn more on mixed fractions here: https://brainly.com/question/414559

#SPJ6

Answer:

A

Step-by-step explanation:

The be the inverse function the domain {4,5,6,7} becomes the range and the range {14,12,10,8} becomes the domain

14 → 4

12 →5

10 →6

8 →7

9514 1404 393

Answer:

  (x, y) = (3, 12)

Step-by-step explanation:

The first step in any problem solving is to look at the problem. Here, we see that the first equation can be reduced to standard form by dividing it by 2. This would give both terms a coefficient of 1.

We see that the second equation already has a variable with a coefficient of 1.

When solving a system by substitution, it can save some effort if you start by finding a variable with a coefficient of 1 or -1. Since we see that in the second equation, we choose to solve the second equation for y:

  y = 4x . . . . . . . add 4x to both sides of the second equation

Now, we have an expression for y that we can substitute into the first equation.

  2x +2(4x) = 30 . . . . substitute for y

  10x = 30 . . . . . . . . simplify

  x = 3 . . . . . . . . . . divde by 10

  y = 4(3) = 12 . . . . . find y using y=4x

The solution is (x, y) = (3, 12).

__

Additional comment

I doesn't matter what variable gets substituted. The purpose of the exercise is to reduce the number of variables in the equation. Here, we start with an equation that has 2 variables. Substituting for one of them gives an equation with only one variable, a lot easier to solve.

You don't have to find an expression for the "bare" variable. We could solve the second equation for 2x, for example: 2x = y/2, then substitute for 2x in the first equation: y/2 +2y = 30   ⇒   y = (2/5)(30) = 12. Or, we could solve the first equation for 2x and substitute into the second: 2x=30-2y, y-2(2x) = 0 ⇒ y-2(30-2y) = 0 ⇒ y = 60/5 = 12.

The substitution property of equality says you can make any substitution of equal values anywhere. "butter = margarine" means you can substitute butter for margarine, or vice versa, wherever you may wish.

Step-by-step explanation:

\underline{\textsf{Given:}}

Given:

\mathsf{Polynomial\;is\;x^3+7x^2+7x-15}Polynomialisx

3

+7x

2

+7x−15

\underline{\textsf{To find:}}

To find:

\mathsf{Factors\;of\;x^3+7x^2+7x-15}Factorsofx

3

+7x

2

+7x−15

\underline{\textsf{Solution:}}

Solution:

\textsf{Factor theorem:}Factor theorem:

\boxed{\mathsf{(x-a)\;is\;a\;factor\;P(x)\;\iff\;P(a)=0}}

(x−a)isafactorP(x)⟺P(a)=0

\mathsf{Let\;P(x)=x^3+7x^2+7x-15}LetP(x)=x

3

+7x

2

+7x−15

\mathsf{Sum\;of\;the\;coefficients=1+7+7-15=0}Sumofthecoefficients=1+7+7−15=0

\therefore\mathsf{(x-1)\;is\;a\;factor\;of\;P(x)}∴(x−1)isafactorofP(x)

\mathsf{When\;x=-3}Whenx=−3

\mathsf{P(-3)=(-3)^3+7(-3)^2+7(-3)-15}P(−3)=(−3)

3

+7(−3)

2

+7(−3)−15

\mathsf{P(-3)=-27+63-21-15}P(−3)=−27+63−21−15

\mathsf{P(-3)=63-63}P(−3)=63−63

\mathsf{P(-3)=0}P(−3)=0

\therefore\mathsf{(x+3)\;is\;a\;factor}∴(x+3)isafactor

\mathsf{When\;x=-5}Whenx=−5

\mathsf{P(-5)=(-5)^3+7(-5)^2+7(-5)-15}P(−5)=(−5)

3

+7(−5)

2

+7(−5)−15

\mathsf{P(-5)=-125+175-35-15}P(−5)=−125+175−35−15

\mathsf{P(-5)=175-175}P(−5)=175−175

\mathsf{P(-5)=0}P(−5)=0

\therefore\mathsf{(x+5)\;is\;a\;factor}∴(x+5)isafactor

\underline{\textsf{Answer:}}

Answer:

\mathsf{x^3+7x^2+7x-15=(x-1)(x+3)(x+5)}x

3

+7x

2

+7x−15=(x−1)(x+3)(x+5)

\underline{\textsf{Find more:}}

Find more:

Answer:

(x-1)^2 + (y+4)^2 = 4

Step-by-step explanation:

The equation for a circle is given by

(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius

(x-1)^2 + (y- -4)^2 = 2^2

(x-1)^2 + (y+4)^2 = 4

Answer:

A. x-2y-3z = 1/2

Step-by-step explanation:

yess.. if you take solve every equation then you will get same equation from all the equations

Answer:

Step-by-step explanation:

The reason you haven't gotten an answer to this is because in its current formatting, there is no answer. Here's what I mean:

The first equation is going to be concerning the NUMBER of memberships sold, where m is male and f is female. The total number of memberships was 10:

m + f = 10

Now for the money equation. The total amount of money made from that number of memberships was 3000, where male memberships cost $300 and so do the female memberships, giving us an equation of

300m + 300f = 3000

Go back to the first equation and solve for either m or f. I solved for m in terms of f:

m = 10 - f and sub that into the second equation for m to get:

300(10 - f) + 300f = 3000 and

3000 - 300f + 300f = 3000. Here is where you find the problem. The -300f and the 300f cancel each other out, leaving you with the fact that

3000 = 3000 which it does, but it doesn't give us any viable answer.

I would have to say that since we can't do math on this, that the most credible answer you'll find is that the same number of male and female memberships were sold because 5 male memberships cost $1500 (5 memberships at $300 a piece is $1500) and 5 female memberships also cost $1500.

1500 + 1500 = 3000

Answer:

A)x>-3

Step-by-step explanation:

as the circle is not coloured this means that -3 is not included so the ones that have

[tex] \geqslant \\ \leqslant [/tex]

are not answers and these means smaller or equal to/greater or equal to.

As the line is going to the right this means that x is greater than -3 so we use > for greater.

so in the end we get that the answer is x > -3

Step-by-step explanation:

Solve for the first variable in one of the equations, then substitute the result into the other equation.

Point Form:

(

12

,

67.45

)

Equation Form:

x

=

12

,

y

=

67.45

Answer:

x-1/2x-3

Step-by-step explanation:

be happy bro this is your perfect answer

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

[tex] \displaystyle \large{ cos \theta = - \frac{1}{2} }[/tex]

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

[tex] \displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }[/tex]

Find Q3

[tex] \displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }[/tex]

Both values are apart of the interval. Hence,

[tex] \displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }[/tex]

Second Question

Isolate sin(4 theta).

[tex] \displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }[/tex]

Rationalize the denominator.

[tex] \displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }[/tex]

The problem here is 4 beside theta. What we are going to do is to expand the interval.

[tex] \displaystyle \large{0 \leqslant \theta < 2 \pi}[/tex]

Multiply whole by 4.

[tex] \displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}[/tex]

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

[tex] \displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }[/tex]

Find Q4

[tex] \displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }[/tex]

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

[tex] \displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}[/tex]

Hence:-

For Q3

[tex] \displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }[/tex]

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

[tex] \displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }[/tex]

Therefore:-

[tex] \displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }[/tex]

Then we divide all these values by 4.

[tex] \displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }[/tex]

Let me know if you have any questions!

Answer:

f = 10

g = 2 sqrt(3)

h = 20

Step-by-step explanation:

The short leg is opposite the smaller angle so it is f

The longer leg is opposite the larger angle so it is g

The hypotenuse is opposite the right angle so it is 20

We know f = x

g = x sqrt(3)

h = 2x = 20

2x = 20  so x = 10

f = 10

g = 2 sqrt(3)

h = 20

Answer:

Because it's a negative.

Step-by-step explanation:

The value of a positive number is still a positive number.

Answer:

7,675

that is your answer

Answer:

(3a - 4 )^2

Step-by-step explanation:

(9a^2-24a+16)=(3a - 4 )^2

Therefore the length of each side of the square is (3a - 4 )^2

Answer:

Part A: (3a - 4)^2

Part B: (5a + 6b)(5a - 6b)

Step-by-step explanation:

Part A:

It looks like this is the square of a binomial. Now we check it.

9a^2 is the square of 3a.

16 is the square of 4 and of -4.

Check the middle term:

2 * 3a * 4 = 24a

2 * 3a * (-4) = -24a

Since we get -24a when we use 4, the second term of the binomial is 4.

Answer:

9a^2 - 24a + 16 = (3a - 4)^2

Part B:

25a^2 − 36b^2

This is a two-term polynomial. The two terms are perfect squares and there is a subtraction sign between them, so this is the difference of two squares. The difference of two squares factors into the product of a sum and a difference.

25a^2 is the square of 5a.

36b^2 is the square of 6b.

25a^2 − 36b^2 = (5a + 6b)(5a - 6b)

Answer:

H0: µd = 0 (claim)

H1: µd ≠ 0

This is a two-tail t-test for µd

Step-by-step explanation:

This is a paired (dependent) sample test, with its hypothesis is written as :

H0: µd = 0

H1: µd ≠ 0

From the equality sign used in the hypothesis declaration, a not equal to ≠ sign in the alternative hypothesis is used for a two tailed t test

The data isn't attached, however bce the test statistic cannot be obtained. However, the test statistic formular for a paired sample is given as :

T = dbar / (Sd/√n)

dbar = mean of the difference ; Sd = standard deviation of the difference.

The pillars and the pyramids in the stadium gate means that we have to calculate the area of the items that make up the gate one after the other. At the end of the calculation, the calculated areas are then added up.

The total cost of painting is Rs.21344

First, we calculate the area of 1 side of 1 pillar using:

[tex]A = Height * Base[/tex]

Where

[tex]Height = 10ft[/tex] --- Height of the pillar

[tex]Base = 3ft[/tex] --- Base of the pillar

So:

[tex]A = 10ft * 3ft[/tex]

[tex]A = 30ft^2[/tex]

The area of the 4 sides of the pillar is:

[tex]A_2 = 4 * A[/tex] --- i.e. 4 multiplied by the area of 1 side

[tex]A_2 = 4 * 30ft^2[/tex]

[tex]A_2 = 120ft^2[/tex]

The area of the 2 pillars is:

[tex]Area_1 = 2 * A_2[/tex] --- i.e. 2 multiplied by the area of 1

[tex]Area_1 = 2 * 120ft^2[/tex]

[tex]Area_1 = 240ft^2[/tex]

Because one part of the pyramid won't be visible, we calculate the area of the pyramid using:

[tex]Area = lw + l\sqrt{(w/2)^2 + h^2} + w\sqrt{(l/2)^2 + h^2}[/tex]

Where:

[tex]h = 2[/tex] -- the height

[tex]l = w = 3[/tex] --- the base of the pillar is the length & width of the pyramid.

So, we have:

[tex]Area = 3\sqrt{(2/2)^2 + 2^2} + 3\sqrt{(2/2)^2 + 2^2}[/tex]

[tex]Area = 3\sqrt{1 + 4} + 3\sqrt{1 + 4}[/tex]

[tex]Area = 3\sqrt{5} + 3\sqrt{5}[/tex]

[tex]Area = 6\sqrt{5}[/tex]

For the two pyramids, the area is:

[tex]Area_2 = 2 * 6\sqrt 5[/tex] -- 2 multiplied by area of 1

[tex]Area_2 = 12\sqrt 5[/tex]

[tex]Area_2 = 26.8[/tex]

So, the total area to be painted is:

[tex]Total = Area_1 + Area_2[/tex] --- the sum of the area of the pillars and the pyramids

[tex]Total = 240+26.8[/tex]

[tex]Total = 266.8ft^2[/tex]

The unit cost of paint is:

Rate = Rs80 per sq.ft

The total cost of painting is:

[tex]Cost = 80 * 266.8[/tex]

[tex]Cost = Rs.21344[/tex]

Hence, the total cost of painting is Rs.21344.

Read more at:

https://brainly.com/question/14320476

Answer:

it is summer how do y'all still have school,like don't they give y'all a break

Answer:

"D"

Step-by-step explanation:

just add the two functions

5x^2 - 8x^2 = -3x^2 etc