Graph the parabola. Plot five points on the parabola. button.​

Answer:

145.8

Step-by-step explanation:

6c+4dc, where c=4.5 d=6.6

6(4.5) + 4(6.6) (4.5)

27 + 118.8

=145.8

Answer:

n=23

Step-by-step explanation:

a right angle measures 90

67+n=90

subtract 67 on both sides

n=23

Use angle addition postulate:

Substitute values to get the required equation and solve it for n:

Answer:

1=32

2=44

3=56

4=70

Step-by-step explanation:

if need working please ask

Depending on the path that we decide to take, the algebra can help us in many forms.

As an example in the pharmaceutical/medical area, the nurses and doctors use basic algebra formulas to calculate dosages on different drugs depending on variables such as the weigh of each patient (commonly expressed as X or Y).

They used to have some paper sheets with formulas for different drug preparations (liquid ones particularly) within hospitals to avoid errors in medication.

As a healthcare provider, it is important to be able to read vital signs. Many of these are expressed as algebraic equations. Such equations can also be important when it comes to administering the right doses of medicine or converting different units of measurement.

Answer:

b 34

Step-by-step explanation:

its 34 because if you do 46×2 because of the both sides you will get 92 then subtract 92 from 160 and you will get 68 then do 68÷2 to find out the second side and you will get 34

Answer:

Option C.

Step-by-step explanation:

Perimeter = 2L + 2h

[tex]P=2(5x^{2}y) +2(3y^{3} )=10x^{2} y+6y^{3}[/tex]

Hope this helps

Answer:

C.When the heart is starved of oxygen

The product of z₁ = 3 · (cos 14π/15 + i · sin 14π/15) and z₂ = 3 √3 · (cos 11π/15 + i · sin 11π/15) in rectangular form with fully simplified expressions is z₁ · z₂ = 7.794 - i · 13.5.

Let be two numbers of the form z = a + i · b, where i = √-1, the product of two of these numbers in rectangular form is described by the following formula:

z₁ · z₂ = (a + i · b) · (c + i · d) = (a · c - b · d) + i · (a · d + b · c)   (1)

If we know that a = 3 · cos 14π/15, b = 3 · sin 14π/15, c = 3√3 · cos 11π/15, d = 3√3 · sin 11π/15, then the result in rectangular form is:

z₁ · z₂ = 7.794 - i · 13.5

The product of z₁ = 3 · (cos 14π/15 + i · sin 14π/15) and z₂ = 3 √3 · (cos 11π/15 + i · sin 11π/15) in rectangular form with fully simplified expressions is z₁ · z₂ = 7.794 - i · 13.5. [tex]\blacksquare[/tex]

The statement presents typing mistakes and is poorly formatted, the correct form is introduced below:

Given the complex number z₁ = 3 · (cos 14π/15 + i · sin 14π/15) and z₂ = 3 √3 · (cos 11π/15 + i · sin 11π/15), express the result of z₁ · z₂ in rectangular form with fully simplified fractions and radicals.

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Explanation:

Any point is of the form (x,y)

The first coordinate is x which tells us to go either left or right depending on whether x is negative or positive.

Example: (-2,3) means we go left 2 units

Another example: (5,7) means we go to the right 5 units.

The starting point is the origin where the x and y axis meet up.

Answer:

False

Step-by-step explanation:

Answer: 11/4

Step-by-step explanation: no need :”)

Answer:

About 5 bags of sand

Step-by-step explanation:

1. Create Equation

[(48x3.14)-(40x3.14)]/6

2. Solve

[(48x3.14)-(40x3.14)]/6

[150.72-125.6]/6

25.12/6

Now if you use a calculater you will get

4.1866666667

So lets just round the number to 5 instead of

4 beacuse if we round it to 4 it won't be enough

to cover the whole path.

so your answer is 5

5. Let [tex]x = \sin(\theta)[/tex]. Note that we want this variable change to be reversible, so we tacitly assume 0 ≤ θ ≤ π/2. Then

[tex]\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2}[/tex]

and [tex]dx = \cos(\theta) \, d\theta[/tex]. So the integral transforms to

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \int \frac{\sin^3(\theta)}{\cos(\theta)} \cos(\theta) \, d\theta = \int \sin^3(\theta) \, d\theta[/tex]

Reduce the power by writing

[tex]\sin^3(\theta) = \sin(\theta) \sin^2(\theta) = \sin(\theta) (1 - \cos^2(\theta))[/tex]

Now let [tex]y = \cos(\theta)[/tex], so that [tex]dy = -\sin(\theta) \, d\theta[/tex]. Then

[tex]\displaystyle \int \sin(\theta) (1-\cos^2(\theta)) \, d\theta = - \int (1-y^2) \, dy = -y + \frac13 y^3 + C[/tex]

Replace the variable to get the antiderivative back in terms of x and we have

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\cos(\theta) + \frac13 \cos^3(\theta) + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\sqrt{1-x^2} + \frac13 \left(\sqrt{1-x^2}\right)^3 + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\frac13 \sqrt{1-x^2} \left(3 - \left(\sqrt{1-x^2}\right)^2\right) + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \boxed{-\frac13 \sqrt{1-x^2} (2+x^2) + C}[/tex]

6. Let [tex]x = 3\tan(\theta)[/tex] and [tex]dx=3\sec^2(\theta)\,d\theta[/tex]. It follows that

[tex]\cos(\theta) = \dfrac1{\sec(\theta)} = \dfrac1{\sqrt{1+\tan^2(\theta)}} = \dfrac3{\sqrt{9+x^2}}[/tex]

since, like in the previous integral, under this reversible variable change we assume -π/2 < θ < π/2. Over this interval, sec(θ) is positive.

Now,

[tex]\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = \int \frac{27\tan^3(\theta)}{\sqrt{9+9\tan^2(\theta)}} 3\sec^2(\theta) \, d\theta = 27 \int \frac{\tan^3(\theta) \sec^2(\theta)}{\sqrt{1+\tan^2(\theta)}} \, d\theta[/tex]

The denominator reduces to

[tex]\sqrt{1+\tan^2(\theta)} = \sqrt{\sec^2(\theta)} = |\sec(\theta)| = \sec(\theta)[/tex]

and so

[tex]\displaystyle 27 \int \tan^3(\theta) \sec(\theta) \, d\theta = 27 \int \frac{\sin^3(\theta)}{\cos^4(\theta)} \, d\theta[/tex]

Rewrite sin³(θ) just like before,

[tex]\displaystyle 27 \int \frac{\sin(\theta) (1-\cos^2(\theta))}{\cos^4(\theta)} \, d\theta[/tex]

and substitute [tex]y=\cos(\theta)[/tex] again to get

[tex]\displaystyle -27 \int \frac{1-y^2}{y^4} \, dy = 27 \int \left(\frac1{y^2} - \frac1{y^4}\right) \, dy = 27 \left(\frac1{3y^3} - \frac1y\right) + C[/tex]

Put everything back in terms of x :

[tex]\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = 9 \left(\frac1{\cos^3(\theta)} - \frac3{\cos(\theta)}\right) + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = 9 \left(\frac{\left(\sqrt{9+x^2}\right)^3}{27} - \sqrt{9+x^2}\right) + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = \boxed{\frac13 \sqrt{9+x^2} (x^2 - 18) + C}[/tex]

2(b). For some constants a, b, c, and d, we have

[tex]\dfrac1{x^2+x^4} = \dfrac1{x^2(1+x^2)} = \boxed{\dfrac ax + \dfrac b{x^2} + \dfrac{cx+d}{x^2+1}}[/tex]

3(a). For some constants a, b, and c,

[tex]\dfrac{x^2+4}{x^3-3x^2+2x} = \dfrac{x^2+4}{x(x-1)(x-2)} = \boxed{\dfrac ax + \dfrac b{x-1} + \dfrac c{x-2}}[/tex]

5(a). For some constants a-f,

[tex]\dfrac{x^5+1}{(x^2-x)(x^4+2x^2+1)} = \dfrac{x^5+1}{x(x-1)(x+1)(x^2+1)^2} \\\\ = \dfrac{x^4 - x^3 + x^2 - x + 1}{x(x-1)(x^2+1)^2} = \boxed{\dfrac ax + \dfrac b{x-1} + \dfrac{cx+d}{x^2+1} + \dfrac{ex+f}{(x^2+1)^2}}[/tex]

where we use the sum-of-5th-powers identity,

[tex]a^5 + b^5 = (a+b) (a^4-a^3b+a^2b^2-ab^3+b^4)[/tex]

Step-by-step explanation:

this is what i get , i hope this will help you

Answer: p=11

Step-by-step explanation:

-3(p-4) = -2p+1

-3p+12=-2p+1

+2p +2p

-p+12=1

-12 -12

(-1)-p=-11 (-1)

p = 11

Check answer:

-3(11-4)=-2(11)+1

-3(7)=(-22)+1

-21 = -21

Answer:

The first one should belong in the 54x+18 due to multiplication

The second answer choice has distributive property which has to belong in the 54x+18 category

The third answer choice fits into the 6 times 9x + 6 times one category due to multiplication like the first answer choice.

The last one should be in the 54x+18 category because of the distributive property

Step-by-step explanation:

The Objectives: 54x+18 and (6(9x) + 6

The first one should belong in the 54x+18 due to multiplication

The second answer choice has distributive property which has to belong in the 54x+18 category

The third answer choice fits into the 6 times 9x + 6 times one category due to multiplication like the first answer choice.

The last one should be in the 54x+18 category because of the distributive property

Answer:

[tex]\sqrt{113}\\[/tex]

Step-by-step explanation:

[tex]distance= \sqrt{(5-[-3])^2+(2-[-5])^2}\\\\distance= \sqrt{(8)^2+(7)^2}\\\\distance= \sqrt{64+49}\\\\distance= \sqrt{113}\\[/tex]

Answer:

Step-by-step explanation:

Given binomial

Use the property

The binomial remains same without brackets

Let the equation be

[tex] \rightarrow y = mx + c[/tex]

[tex] \huge\rightarrow \: m = \frac{y' - y}{x' - x} \\ [/tex]

[tex] \huge\rightarrow \: m = \frac{4 - 0 }{0 \: - 1} = - 4 \\ [/tex]

hence ,our equation would be,

[tex]\huge \rightarrow 4x + y = 4[/tex]

Base on the equidistant chords theorem, chords FB and DA are congruent because they are equidistant from C.

The equidistant chords theorem states that two chords in the same circle are congruent to each other, if they are equidistant from the center of the circle.

Thus, based on the information given we can conclude that chords FB and DA are congruent because they are equidistant from C based on the equidistant chords theorem.

Learn more about the equidistant chords theorem on:

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Answer:

Its D The one at the top is incorrect.

Step-by-step explanation:

I got A 100% which means i got it right...

Assuming  He wants to earn interest on this money rather than keep extra money he does not need. The choice that would be the best for his situation is  $250 to open a regular savings.

A regular savings account is a saving acount that enables you to save and earn interest based on the amount saved because it is an interest bearing account.

Although this account gives you interest but the interest rate is low. Since Tim has $1200 in his checking account  assuming he withdraw $100 twices, Tim will have $1,000 ($1,200-$200).

Now if he open a regular saving with $250 he will have $750 left ($1,000-$250) with which he can earn interest with.

Inconclusion the choice that would be the best for his situation is  $250 to open a regular savings.

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[tex]x = vt\\\\\implies x= (x+4t)t\\\\\implies x=xt+4t^2\\\\\implies x -xt= 4t^2\\\\\implies x (1-t) =4t^2\\\\\implies x = \dfrac{4t^2}{1-t}[/tex]

Answer:

positive, leaner association

Step-by-step explanation:

Answer:

The first thing to do is translate the sentence into mathematical notation.

"no more than" means "less than or equal to" which is written as ≤

"seven less than a number" means we are subtracting 7 from a number. We don't know what the number is, so we can use a variable (like n) for the number. So "seven less than a number" becomes n - 7.

The whole thing becomes 12 ≤ n - 7.

To solve for n, we can add 7 to both sides.

19 ≤ n.

19 is less than or equal to n, which means the smallest (minimum) value of the number is 19.

We could also do it without an inequality:

If 12 is 7 less than a number, then the number is 19, because 19-7=12. If we chose a smaller number, like 18, then 7 less than 18 is smaller than 12, which is not allowed (12 is no more than 7 less than the number). So the smallest possible value of the number is 19.

Let the number be p.

Next, "seven times p" can be written like so:-

[tex]\pmb{7p}[/tex]

This expression is less than 18:-

[tex]\bigstar{\boxed{\pmb{7p < 18}}}[/tex]

Hope everything is clear; if you need any explanation/clarification, kindly let me know, and I will comment and/or edit my answer :)

Answer:

[tex]( \frac{x}{3} ) {}^{2} + ( \frac{y}{3}) {}^{2} = 1[/tex]

or x² +y²= 9

Step-by-step explanation:

In Cartesian form, the equation is expressed only in terms of y and x.

x= -3sin(t) -----(1)

y= 3cos(t) -----(2)

I've written x instead of x(t) as in the later part of the working, we will be having an equation of only x and y, thus x will no longer be a function of t. This applies to equation 2, where I have replaced y(t) with y.

Relating sine to cosine:

sin²(t) +cos²(t)= 1

[sin(t)]² +[cos(t)]²= 1 -----(3)

From (1):

[tex] \frac{x}{ - 3} = \sin(t) [/tex]

[tex] \sin(t) = - \frac{x}{3} [/tex] -----(3)

From (2):

[tex] \frac{y}{3} = \cos(t) [/tex]

[tex] \cos(t) = \frac{y}{3} [/tex] -----(4)

Substitute (4) &(5) into (3):

[tex]( - \frac{x}{3} )^{2} + ( \frac{y}{3}) {}^{2} = 1[/tex]

[tex]( \frac{x}{3} ) {}^{2} + ( \frac{y}{3} ) {}^{2} = 1[/tex]

The steps below are optional as the above is already considered to be the Cartesian form.

[tex] \frac{ {x}^{2} }{9} + \frac{y {}^{2} }{9} = 1[/tex]

Multiplying both sides by 9:

x² +y²= 9

Answer:

  (x, y) = (3, 8)

Step-by-step explanation:

The graph for y is a translation upward 4 units of the graph of f(x).

__

A point on the graph of f(x) is given as ...

  (x, f(x)) = (3, f(3)) = (3, 4)

Then a point on the graph of y will be ...

  (x, y) = (x, f(x) +4)) = (3, f(3)+4) = (3, 4 +4)

  (x, y) = (3, 8)

Answer: Angles 6 and 7

Step-by-step explanation:

These angles form a linear pair, and angles that form a linear pair are supplementary.

Based on the information provided about the return trip, the train's speed on its trip south was 6.36 miles per hour.

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