Problem 5
The sum of the first ten terms of an arithmetic progression consisting of
positive integer terms is equal to the sum of the 20th, 21st and 22nd term.
If the first term is less than 20, find how many terms are required to give
a sum of 960.

Answer:

Step-by-step explanation:

Remark

I have to assume that you know calculus. It is the only way the problem can be done that I know of. If you don't, I'm not sure how you will do this.

The curve is of y  = e^(-2x) + x^2 - 3

The curve crosses the y axis when x = 0. The y value is

y = e^0 + x^2 - 3

yint = 1 + 0 - 3

yint = -2

The slope at point (0,-2) is

y' = -2e^(-2x) +2x

y' = -2 at point A

Therefore the normal will have a slope

m1 * m2 = - 1

The slope of the curve C  at A = -2

The equation of the tangent line at A = -2x - 2

Call this m1

m2 = slope of the normal

-2 * m2 = -1

m2 = 1/2

Equation of the line (l) =

y = 1/2 x  - 2

The graph is shown below. Notice the two lines actually look like they are at a 90 degree angle.

Answer:

D 14

Step-by-step explanation:

Let x be the number

2(x-4) = 20

Divide each side by 2

2/2(x-4) = 20/2

x-4 = 10

Add 4 to each side

x-4+4 = 10+4

x = 14

Answer:

The number is 14

Step-by-step explanation:

Let's call this unknown number of x:

2×(x – 4) = 20

2x – 8 = 20

2x = 20 + 8

2x = 28

x = 28/2

x = 14

Answer:

Step-by-step explanation:

Step-by-step explanation:

hope this helps to u......

Answer:

£10.1

Step-by-step explanation:

Highest exchange rate=1.3401

Lowest exchange rate=1.3199

The money Khan would have received more if he exchanged currencies at the highest rate compared to the lowest rate=500*(1.3401-1.3199)=£10.1

Answer:

x²+ x = 0

x ( x +1 ) = 0

x = -1

x³-25x=0

x ( x² - 25 ) = 0

x² = 25

x = 5

x²=4x

x² - 4x = 0

x ( x- 4 ) = 0

x = 4

4x²-4x=1

4x²-4x - 1 = 0

4 x² - 2x - 2x - 1 = 0

2x ( 2x - 1 ) - 1 ( 2x- 1 ) = 0

2x- 1 ) ² = 0

2x = 1

x = 1/ 2

mark me as brainliest

Answer:

=X-2a

= 6- 2× (-7)

= 6 + 14

= 20

✌✌✌✌

Answer:

right by 3 I think

Step-by-step explanation:

Answer: Right by 3 Units

Step-by-step explanation:

Right on edge 2021

Answer:

2^3 x 31

Step-by-step explanation:

31 is a prime number

Answer: 4. A base area of 16y^2 square units and height of y^2 + y + 3 units

Step-by-step explanation:

Using the distributive property; you can see that 16y^2(y^2+y+3)=

16y^4+16y^3+48y^2

Answer:

D. a base area of 16y2 square units and height of y^2 + y + 3 units

Step-by-step explanation:

Ed22

Answer:

The markup is 81.25%

Step-by-step explanation:

Take the new price and subtract the original price

58 - 32

26

Divide by the original price

26/32

.8125

Change to percent form

81.25%

The markup is 81.25%

Answer:

Hello,

Step-by-step explanation:

[tex]\dfrac{x^{\frac{5}{6}} }{x^{\frac{1}{6}} } \\\\=x^{\frac{5}{6} -\frac{1}{6} }\\\\=x^{\frac{4}{6} }\\\\=x^{\frac{2}{3} }\\\\=\sqrt[3]{x^2}[/tex]

Hi there!

Look the picture for your answer.

Hope it helps!

Answer:

Centre (0,0)

Radius √9 = 3

I've attached a graph if the circle with it

Answered by GAUTHMATH

Answer:

First, we apply the Distributive property and then we combine like terms,

To combine like terms, we add or subtract.

[tex]-(2x +y) - 2 ( -x - y)[/tex]

[tex]=-2x-y+2x+2y[/tex]

[tex]=(-2x+2x)+(-y+2y)[/tex]

[tex]=0+y[/tex]

[tex]=y[/tex]

OAmalOHopeO

Answer:

y is the simplest result here.

Step-by-step explanation:

Perform the indicated multiplication as a first step towards simplifying this expression:

-2x - y + 2x + 2y

-2x and 2x cancel each other out, leaving 2y - y, or just y

Answer:-24(x-3) = -24x (-24)(-3)

Step-by-step explanation:

Answer:

eat my hand where the pic

Step-by-step explanation:

yeah

Answer:

SAS

43ft

Step-by-step explanation:

We know that two sides are equal

PQ = ST and QR = TU and the angle between them is equal Q = T

We can use the SAS (side angle side)

Since the triangles are congruent

PR = SU

6y+5 = 8y

Subtract 6y from each side

6y+5-6y = 8y-6y

5 = 2y

Divide by 2

5/2 = y

2.5 = y

PR = ST = 8y = 8(2.5) =20

The perimeter is 9+14+20 = 43

Answer:

B. =35ft

Step-by-step explanation:

The perimeter of a right angled triangle is a+b+c

which is 9+4+6y+5 =90

collect like terms

9+4+5+6y=90

18 + 6y = 90

6y = 90 - 18

6y= 72

divide both sides by 6

6y/6 = 72/6

y = 12

so therefore, the perimeter of ∆PQR is

9 + 12 + 14

= 35ft

Because all the angles are congruent (the same), this is an equilateral triangle. All equilateral triangles have congruent angles and congruent sides, so all sides has to be 14.

Answer:

cosA = 0.6

Step-by-step explanation:

Using the Pythagorean identity

sin²A + cos²A = 1 ( subtract sin²A from both sides )

cos²A = 1 - sin²A ( take the square root of both sides )

cosA = ± [tex]\sqrt{1-sin^2A[/tex]

Since only the positive value is required , then

cosA = [tex]\sqrt{1-(0.8)^2}[/tex]

         = [tex]\sqrt{1-0.64}[/tex]

         = [tex]\sqrt{0.36}[/tex]

         = 0.6

Answer:

Answer:

Answer:x = 25°

Answer:x = 25°Step-by-step explanation:

Answer:x = 25°Step-by-step explanation:The secant- secant angle FGE is one half the difference of the intercepted arcs, that is

Answer:x = 25°Step-by-step explanation:The secant- secant angle FGE is one half the difference of the intercepted arcs, that is\frac{1}{2}

Answer:x = 25°Step-by-step explanation:The secant- secant angle FGE is one half the difference of the intercepted arcs, that is\frac{1}{2} 2

Answer:x = 25°Step-by-step explanation:The secant- secant angle FGE is one half the difference of the intercepted arcs, that is\frac{1}{2} 21

Answer:x = 25°Step-by-step explanation:The secant- secant angle FGE is one half the difference of the intercepted arcs, that is\frac{1}{2} 21

Answer:x = 25°Step-by-step explanation:The secant- secant angle FGE is one half the difference of the intercepted arcs, that is\frac{1}{2} 21 ( BD - FE ) = ∠ FGE, that is

Answer:x = 25°Step-by-step explanation:The secant- secant angle FGE is one half the difference of the intercepted arcs, that is\frac{1}{2} 21 ( BD - FE ) = ∠ FGE, that is\frac{1}{2}

Answer:x = 25°Step-by-step explanation:The secant- secant angle FGE is one half the difference of the intercepted arcs, that is\frac{1}{2} 21 ( BD - FE ) = ∠ FGE, that is\frac{1}{2} 2

Answer:x = 25°Step-by-step explanation:The secant- secant angle FGE is one half the difference of the intercepted arcs, that is\frac{1}{2} 21 ( BD - FE ) = ∠ FGE, that is\frac{1}{2} 21

Answer:x = 25°Step-by-step explanation:The secant- secant angle FGE is one half the difference of the intercepted arcs, that is\frac{1}{2} 21 ( BD - FE ) = ∠ FGE, that is\frac{1}{2} 21

Answer:x = 25°Step-by-step explanation:The secant- secant angle FGE is one half the difference of the intercepted arcs, that is\frac{1}{2} 21 ( BD - FE ) = ∠ FGE, that is\frac{1}{2} 21 (135 - x) = 55 ( multiply both sides by 2 to clear the fraction )

Answer:x = 25°Step-by-step explanation:The secant- secant angle FGE is one half the difference of the intercepted arcs, that is\frac{1}{2} 21 ( BD - FE ) = ∠ FGE, that is\frac{1}{2} 21 (135 - x) = 55 ( multiply both sides by 2 to clear the fraction )135 - x = 110 ( subtract 135 from both sides )

Answer:x = 25°Step-by-step explanation:The secant- secant angle FGE is one half the difference of the intercepted arcs, that is\frac{1}{2} 21 ( BD - FE ) = ∠ FGE, that is\frac{1}{2} 21 (135 - x) = 55 ( multiply both sides by 2 to clear the fraction )135 - x = 110 ( subtract 135 from both sides )- x = - 25 ( multiply both sides by - 1 )

Answer:x = 25°Step-by-step explanation:The secant- secant angle FGE is one half the difference of the intercepted arcs, that is\frac{1}{2} 21 ( BD - FE ) = ∠ FGE, that is\frac{1}{2} 21 (135 - x) = 55 ( multiply both sides by 2 to clear the fraction )135 - x = 110 ( subtract 135 from both sides )- x = - 25 ( multiply both sides by - 1 )x = 25°

Step-by-step explanation:

A

9×h - 2

B

the baseline of the triangle is 2 units less than 9 times the height long.

C

h = 1/3

9×1/3 - 2 = 3 - 2 = 1

Answer:

A, B, B

Step-by-step explanation:

Answer:

b

Step-by-step explanation:

cuz u add them all

Answer:

210

Step-by-step explanation:

In a normal distribtuion mean=mode=median

so 210=median

Answer:

5th option

Step-by-step explanation:

The secant- tangent angle DCB is half the difference of the intercepted arcs, that is

∠ 1 = [tex]\frac{1}{2}[/tex] ( AD - BD)

Answer:

surface area of a square = 2 ( lb + bh + hl )

given that,

length = 8cm

breadth = 10cm

height = 15 cm

surface area = 2 ( 8* 10 + 10*15 + 15*8 )

                     = 2 ( 80 + 150 + 120 )

                     = 2 * 350

                     = 700 [tex]cm^{2}[/tex]

hope this answer helps you!!

A general cosine function (we could also use a sine function) is written as:

y = A*cos(k*x + p) + M

We will find that the function of the graph is:

f(x) = 2*cos(2*x + 2.09) - 2

Let's return to the general function:

y = A*cos(k*x + p) + M

A is the amplitude, it defines the distance between the value of a maximum and the value of the minimum, such that A is exactly half of that difference.

Here we can see that the maximum is 0, and the minimum is -4

The differene is: 0 - (-4) = 4

Then:

A = 4/2 = 2

f(x) = 2*cos(k*x + p) + M.

M is the midline, this is, the horizontal line that cuts the graph in two halves. Here we can see that the midline is x = -2, then:

M = -2

f(x) = 2*cos(k*x + p) - 2

p is the phase shift.

In the graph, we can see that f(0) = -3, so we have:

f(0) = 2*cos(0 + p) - 2 = -3

      cos(p) = -1/2

             p = Acos(-1/2) = 2.09

Then we have:

f(x) = 2*cos(k*x + 2.09) - 2

Finally, k is related to the frequency of the function.

We can see that the function does a complete cycle at x = pi

This means that:

f(x) = f(x + pi)

Knowing that the period of a cosine function is 2*pi, then:

k*(x + pi) = k*x + 2*pi

k = 2

Then the equation of the graph is:

f(x) = 2*cos(2*x + 2.09) - 2

If you want to learn more, you can read:

https://brainly.com/question/24372261

Answer(s):

[tex]\displaystyle y = 2sin\:(2x + 1\frac{1}{4}\pi) - 2 \\ y = 2cos\:(2x - 1\frac{1}{4}\pi) - 2[/tex]

Step-by-step explanation:

[tex]\displaystyle y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow -2 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-\frac{5}{8}\pi} \hookrightarrow \frac{-1\frac{1}{4}\pi}{2} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\pi} \hookrightarrow \frac{2}{2}\pi \\ Amplitude \hookrightarrow 2[/tex]

OR

[tex]\displaystyle y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow -2 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{\frac{5}{8}\pi} \hookrightarrow \frac{1\frac{1}{4}\pi}{2} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\pi} \hookrightarrow \frac{2}{2}\pi \\ Amplitude \hookrightarrow 2[/tex]

You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the cosine graph, if you plan on writing your equation as a function of sine, then by all means, go for it, but be careful and follow what is explained here. Now, as you can see, the photograph on the right displays the trigonometric graph of [tex]\displaystyle y = 2sin\:(2x - 1\frac{1}{4}\pi) - 2,[/tex] in which you need to replase “cosine” with “sine”, then figure out the appropriate C-term that will make the graph horisontally shift and map onto the cosine graph [photograph on the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the sine graph [photograph on the right] is shifted [tex]\displaystyle \frac{\pi}{4}\:unit[/tex] to the right, which means that in order to match the cosine graph [photograph on the left], we need to shift the graph BACKWARD [tex]\displaystyle \frac{\pi}{4}\:unit,[/tex] which means the C-term will be negative. Now, BEFORE we go any further, we must remember that this particular cosine graph [thank goodness it is a cosine graph we are working with] ALREADY has a horisontal shift and does not have a single crest oscıllαtıng about any endpoint on the y-axis. So, in this case we need to figure out how far the FIRST oscıllαtıng crest is from the origin, and that obviously would be [tex]\displaystyle \frac{5}{8}\pi\:units.[/tex] Though, sinse we want the sine equation of this graph, it must be “negative”; so, by perfourming your calculations, you will arrive at [tex]\displaystyle \boxed{-\frac{5}{8}\pi} = \frac{-1\frac{1}{4}\pi}{2},[/tex] in which the value of C is [tex]\displaystyle -1\frac{1}{4}\pi.[/tex] So, the sine equation of the cosine graph, accourding to the horisontal shift, is [tex]\displaystyle y = 2sin\:(2x + 1\frac{1}{4}\pi) - 2.[/tex] Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph hits [tex]\displaystyle [\frac{7}{8}\pi, -2],[/tex] from there to [tex]\displaystyle [-\frac{\pi}{8}, -2],[/tex] they are obviously [tex]\displaystyle \pi\:units[/tex] apart, telling you that the period of the graph is [tex]\displaystyle \pi.[/tex] Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the midline. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at [tex]\displaystyle y = -2,[/tex] in which each crest is extended two units beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

**As you can see, this is one of those moments where you will really need to be careful because if you notised, both equations have OPPOCITE horisontal shifts and C-values. Now, the ONLY TIME this occurs is when all crests in a SINUSOIDAL graph cycle half-way in between endpoints. Your best bet is to jot this down for when you see graphs like these in the future.

I am delighted to assist you at any time.