Find the maturity value of a loan of ​$800 after three years. The loan carries a simple interest rate of 9 1/4​% per year.

1.) The maturity of the loan is $___

Answer:

The distance is 20 units

Step-by-step explanation:

Step-by-step explanation:

A = 0.5(πr^2 )

A = 0.5(π × (20^2 ))

A = 628.32 units^2

Answer:

628.3 ft

Step-by-step explanation:

area of half circle = [tex]\frac{\pi r^{2}}{2}[/tex]

                             = [tex]\frac{\pi *20^{2} }{2}[/tex]

                             = 200[tex]\pi[/tex] = 628.3 ft

Answer:

113.04cm^2

Step-by-step explanation:

The area of a circle is pi * r^2

pi=3.14

So 6^2 is 36

36*3.14=113.04

Hope it helps!

The apportion of Bronx quota is the alloted profit to Bronx

The apportion for Bronx quota is 20

The dataset is given as:

From the dataset, we have:

Bronx: 20.73

Remove the numbers after the decimal point

Bronx =  20

Hence, the apportion for Bronx quota is 20

Read more about quota at:

https://brainly.com/question/1499498

The area of the shaded part of the given figure is 60 square inches'

The given figure is made up of a rectangle and 2 similar triangles

Area of the shaded part = Area of rectangle + area of 2 triangles

From the figure

Area of rectangle = 12 * 4 = 48 square in

Area of one triangle = 0.5(6)(4) = 12 square in

Area  of the figure = 48 + 12

Area  of the figure = 60 square inches'

Hence the area of the shaded part is 60 square inches.

Learn more on area of composite figure here: https://brainly.com/question/15981553

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]

Answer:

-60p -6

Step-by-step explanation:

8(1 - 6p) - 2(6p + 7)

Distribute

8 - 48p  -12p -14

Combine like terms

-60p -6

1320

Step-by-step explanation:

the triangle,1/2×15×20×2=300 .rectangle,17×25=425,20×17=340,15×17=255. sum them up 300+255+425+340=1320

Step-by-step explanation:

Given :-

To find:-

Solution :-

Formula for perimeter of rectangular swamp = 2(L+W)

124 ft. = 2( 5x -10 + x )

124 ft. = 2(6x - 10)

124 ft. = 12x - 20

124+20 = 12x

144/12 = x

12 = x

putting the values of x

Length = 5x-10

width = X

perimeter =2L+2W or 2(L+W)

124= 2(5W-10+W)

perimeter =2( 6W-10)

124 = 12W-20

124+20=(12W+20-20

144=12W

144÷12=12W÷12

12=W

Answer:

Its 9.0 units

Step-by-step explanation:

For the side mostly to the left, its gonna be 3 units, since we can easlily count it

Now, for the diagonals, we have to use the formula for the area of a triangle, which is (l * w) / 2 = a.

For the bottom side, its 3 along the x axis and 1 along the y axis. 3 * 1 is 3, now divide that by 2 and you got 1.5

Now, for the right side, its 5 on the y, 1 on the x. Multiply 1 by 5, you got 5. Now divide that by 2 and you have 2.5

Finally, for the top side, its 4 on the x, and 1 on the y. 4 * 1 is 4, divided by 2 and its 2.

Alright, now we have to add 'em all together. 3 + 1.5 + 2.5 + 2 is 9 units

Theres your answer :)

Area of the yellow sector is 58.91 cm2

The spinner has a circular shape and would be solved as a circle.

Area of a circle = pi r^2

= 3.142 * 5^2

= 78.55 cm2

The circle is divided into 8

78.55 / 8 = 9.8188

Area of 6 parts = 6 * 9.8188
= 58.9128 cm2

Answer:

7k + 6

Steps:

Subtract the numbers 20+6k-14k+k

6+6k+k

Combine the terms:

6+6k+k

6+7k

Rearange terms:

6+7k

7k+6

bought price= 2000

sold price= 2500

the percentage profit

[tex]profit\% = 100 \times \frac{(final - initial)}{initial} [/tex]

[tex]profit\% = 100 \times \frac{(2500 - 2000)}{2000} [/tex]

[tex] = \frac{(2500 - 2000)}{2000} [/tex]

[tex] = \frac{1}{4} [/tex]

[tex] = \frac{1}{4} \times 100\%[/tex]

[tex]profit\% = 25\%[/tex]

therefore, his percentage profit is 25.

Answer: Its negetive. The answer is -52d^2

Hope this helped you!

Answer:

Step-by-step explanation:

[tex]-12\ d^{2}+-40\ d^{2} = -12\ d^{2}-\ 40\ d^{2} = -52\ d^{2}[/tex]

Since −52 is negative and d² is always positive for any number d

Then ,their product −52 d² is negative .

Answer:

it is very high because

Step-by-step explanation:

5-2+3 so maybe it a subtraction problem hope this works

Answer:

r=14/2

r=7

[tex]circumference = 2\pi {r}\\ = 2 \times \frac{22}{7} \times 7 \\ = 44[/tex]

[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]

Let's find the circumference of given circle ~

Circumference :

[tex] \qquad \tt  \dashrightarrow \: \pi{d}^{} [/tex]

where, d = diameter

Now, let's calculate it ~

[tex] \qquad \tt  \dashrightarrow \: c = 14\pi{}^{} [/tex]

That's the answer I terms of π, but we can also plug in the approximate value of pi as 3.14

[tex] \qquad \tt  \dashrightarrow \: c = 14 \times 3.14{}^{} [/tex]

[tex] \qquad \tt  \dashrightarrow \: c = 43.96 \: \: m[/tex]

[tex] \qquad \tt  \dashrightarrow \: c \approx 44 \: \: m[/tex]

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:

The answer is 340 ft²

To find the total area we make two squares and a big rectangle.

1st square

3x5=15

2nd square

5x5=25

Rectangle

20x15=300

Total Area

15+25+300 = 340ft²

Answer:

When you graph the equations, both equations represent the same line. If a system has no solution, it is said to be inconsistent . The graphs of the lines do not intersect, so the graphs are parallel and there is no solution.

Base on the equation, the amount Laura pay for 2 tennis balls and 4 golf balls $8

Sam buys 4 tennis balls and 12 golf balls for a total of £19.

let

x = price of each tennis ball

y = price of each golf ball

Hence,

4x + 12y = 19

Luke buys 6 tennis ball and 10 golf balls for a total of £22.50. Hence,

6x + 10y = 22.50

Combine the equation

4x + 12y = 19

6x + 10y = 22.50

6x + 18y = 28.5

6x + 10y = 22.50

8y = 6

y = 6 / 8

y = $0.75

6x + 10(0.75) = 22.50

6x + 7.5 = 22.50

6x = 22.50 - 7.5

6x = 15

x = 15 / 6

x = $2.5

The cost when Laura pays for 2 tennis balls and 4 golf balls is as follows:

2(2.5) + 4(0.75) = 5 + 3 = $8

learn more on equation here: https://brainly.com/question/17290525

Answer:

A. X < 5

Step-by-step explanation:

6(x -2) + 21 < 39

6x - 12 + 21 < 39

Group like terms

6x < 39 + 12 - 21

6x < 30

Divide through by 6

X < 30/6

X < 5

Answer:

$25.40

Step-by-step explanation:

Find the unit price. So what you do is just divide 304.80 by 12 to get the price of one which comes out to be $25.40

Answer:

3/4

Step-by-step explanation:

Give brainliest if it helped...(:

Answer:

The answer is $1.85

Step-by-step explanation:

Now, we know that the cost of 5 chocolate bars are $5.75.

For 1 chocolate bar

5.75 ÷ 5 = $1.15

Here, we get the price of 1 chocolate bar

So, the cost of 2 chocolate bars and 3 packets of sweets are $7.85.

2 × chocolate bar = 2 × (1.15) = $2.3

Now, we want to find only the cost of one packet of sweets.

So,

7.85 – 2.3 = $5.55

3 packets of sweets cost $5.55

For 1 packets of sweets

5.55 ÷ 3 = $1.85

Thus, The cost of one packet of sweets = $1.85

Answer:

I can help!! answer in image

Step-by-step explanation:

Have a great day sir

Answer:

x is equal to -34/16

Step-by-step explanation:

Hey there!

In order to solve for x, we need to first simplify everything

After simplifying, it will look like this:

[tex](4x^2+14x+16)-7x+14=4x^2-4-9x[/tex]

Now you can simplify by combining like terms

[tex]4x^2+7x+30=4x^2-9x-4[/tex]

You can cross out [tex]4x^2[/tex] from both sides and then simplify even more by again combining like terms

[tex]16x+34=0\\16x=-34\\x=-34/16[/tex]

So x is equal to -34/16

To solve the given equation, let's simplify and expand both sides:

[tex]\displaystyle\sf (2x+1)(2x+6)-7(x-2)=4(x+1)(x-1)-9x[/tex]

Expanding the brackets:

[tex]\displaystyle\sf (4x^{2}+13x+6)-7x+14=4(x^{2}-1)-9x[/tex]

Simplifying further:

[tex]\displaystyle\sf 4x^{2}+13x+6-7x+14=4x^{2}-4-9x[/tex]

Combining like terms:

[tex]\displaystyle\sf 4x^{2}+6x+20=4x^{2}-9x-4[/tex]

Subtracting [tex]\displaystyle\sf 4x^{2}[/tex] from both sides:

[tex]\displaystyle\sf 6x+20=-9x-4[/tex]

Bringing all the variables to one side and the constants to the other side:

[tex]\displaystyle\sf 6x+9x=-4-20[/tex]

[tex]\displaystyle\sf 15x=-24[/tex]

Dividing both sides by [tex]\displaystyle\sf 15[/tex] to solve for [tex]\displaystyle\sf x[/tex]:

[tex]\displaystyle\sf x=-\frac{24}{15}[/tex]

Simplifying the fraction:

[tex]\displaystyle\sf x=-\frac{8}{5}[/tex]

Therefore, the solution to the equation is [tex]\displaystyle\sf x=-\frac{8}{5}[/tex].

[tex]\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}[/tex]

♥️ [tex]\large{\underline{\textcolor{red}{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}[/tex]