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3 years ago

1) You are designing a new kiddie soccer arena. The area of the rectangular

Mathematics

1 answer:

nikdorinn [45]3 years ago

7 0

Answer:

The length of the rectangular field is 12 ft and the width is 5 ft.

Step-by-step explanation:

The area of a rectangle is:

$\text{Area} = \text{length}\times \text{width}$

The area of a rectangular field is, <em>A</em> = 60 ft.

The length of the field is:

l = w + 7

Compute the width of the field as follows:

$\text{Area} = \text{length}\times \text{width}$

$60=(w+7)\times w\\\\60=w^{2}+7w\\\\w^{2}+7w-60=0\\$

Factorize the last equation by splitting the middle term as follows:

$w^{2}+7w-60=0\\w^{2}+12w-5w-60=0\\w(w+12)-5(w+12)=0\\(w-5)(w+12)=0$

Now, since the width of rectangular field cannot be negative, the width is 5 ft.

Compute the length as follows:

l = w + 7

= 5 + 7

= 12

The length of the rectangular field is 12 ft.

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You deposit $300 in a savings account that pays 6% interest compounded semiannually. How much will you have at the middle of the

Makovka662 [10]

Answer:

The total amount accrued, principal plus interest, from compound interest on an original principal of $ 300.00 at a rate of 6% per year compounded 2 times per year over 0.5 years is $ 309.00.

The total amount accrued, principal plus interest, from compound interest on an original principal of $ 300.00 at a rate of 6% per year compounded 2 times per year over 1 year is $ 318.27.

Step-by-step explanation:

a) How much will you have at the middle of the first year?

Using the formula

$A\:=\:P\left(1+\frac{r}{n}\right)^{nt}$

where

Principle = P

Annual rate = r

Compound = n

Time = (t in years)

A = Total amount

Given:

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 0.5 years

To determine:

Total amount = A = ?

Using the formula

$A\:=\:P\left(1+\frac{r}{n}\right)^{nt}$

substituting the values

$A=300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(0.5\right)}$

$A=300\cdot \frac{2.06}{2}$

$A=\frac{618}{2}$

$A=309$ $

Therefore, the total amount accrued, principal plus interest, from compound interest on an original principal of $ 300.00 at a rate of 6% per year compounded 2 times per year over 0.5 years is $ 309.00.

Part b) How much at the end of one year?

Using the formula

$A\:=\:P\left(1+\frac{r}{n}\right)^{nt}$

where

Principle = P

Annual rate = r

Compound = n

Time = (t in years)

A = Total amount

Given:

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 1 years

To determine:

Total amount = A = ?

so using the formula

$A\:=\:P\left(1+\frac{r}{n}\right)^{nt}$

so substituting the values

$A\:=\:300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(1\right)}$

$A=300\cdot \frac{2.06^2}{2^2}$

$A=318.27$ $

Therefore, the total amount accrued, principal plus interest, from compound interest on an original principal of $ 300.00 at a rate of 6% per year compounded 2 times per year over 1 year is $ 318.27.

3 0

3 years ago

Evaluate the function below for x = 4.<br> F(x) = x3 + 2x2 + 1

pav-90 [236]

Answer:

Step-by-step explanation:

Plug in:

F(4) = (4)3 + 2(4)2 + 1

F(4) = 12 + (8)^2 + 1

F(4) = 12 + 64 + 1

F(4) = 77

(assuming 2x2 = 2x squared)

4 0

3 years ago

You are designing a rectangular poster to contain 7575 in2 of printing with a 33-in margin at the top and bottom and a 11-in m

erma4kov [3.2K]

Answer:

The dimensions of the rectangular poster is 15 in by 5 in.

Step-by-step explanation:

Given that, the area of the rectangular poster is 75 in².

Let the length of the rectangular poster be x and the width of the rectangular poster be y.

The area of the poster = xy in².

$\therefore xy=75$

$\Rightarrow y=\frac{75}{x}$ ....(1)

1 in margin at each sides and 3 in margin at top and bottom.

Then the length of printing space is= (x-2.3) in

=(x-6) in

The width of printing space is = (y-2.1) in

=(y-2) in

The area of the printing space is A =(x-6)(y-2) in²

∴ A =(x-6)(y-2)

Putting the value of y

$\Rightarrow A =(x-6)(\frac{75}{x}-2)$

$\Rightarrow A = 87-\frac{450}{x}-2x$

Differentiating with respect to x

$A '= \frac{450}{x^2}-2$

Again differentiating with respect to x

$A''=-\frac{900}{x^3}$

To find the minimum area of printing space, we set A' = 0

$\therefore \frac{450}{x^2}-2=0$

$\Rightarrow 450 =2x^2$

$\Rightarrow x^2=225$

$\Rightarrow x=\pm 15$

Now putting x=±15 in A''

$A''|_{x=15}=-\frac{900}{15^3}$

$A''|_{x=-15}=-\frac{900}{(-15)^3}=\frac{900}{(15)^3}>0$

Since at x=15 , A"<0 Therefore at x=15 , the area will be minimize.

From (1) we get

$y=\frac{75}{x}$

Putting the value of x

$y=\frac{75}{15}$

=5 in

The dimensions of the rectangular poster is 15 in by 5 in.

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3 years ago

What is the ratio 15 girls to 6 boys in simplest form?

Alenkinab [10]

5 girls to 2 boys, it’s not asking for boys and girls

4 0

3 years ago

Read 2 more answers

what is the y-intercept of the equation of the line that is perpendicular to the line y = 3/5x 10 and passes through the point (

bazaltina [42]

Perpendiculare means the slopes multiply to -1

y=3/5x+10
slope is 3/5
3/5 times what is -1
-5/3 is answer

slope of perpenduclar is-5/3 find y int
y=-5/3+b
point given is (15,-5)
x=15
y=-5

-5=-5/3(15)+b
-5=-25+b
add 25 to both sides
20=b

the yintercept is y=20 or the oint (0,20)

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3 years ago