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2 years ago

14

Charles takes a loan of $15000 on a 7% interest rate. He plans to pay the loan off in full after 2 years. Calculate the simple i

nterest.

1 answer:

Sladkaya [172]2 years ago

5 0

Answer:

$2100

Step-by-step explanation:

p x r x t

15000 x 0.07 x 2

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At the city museum, child admission is $5.20 and adult admission is $9.50. On Friday,133 tickets were sold for a total sales of

fredd [130]

Answer:

55 adult tickets were sold on Friday.

Step-by-step explanation:

A linear system will be used to model the system.

Let c be the number of child tickets and a be the number of adult tickets

Then according to given statements

$a+c = 133\ \ \ \ Eqn\ 1\\9.50a+5.20c = 928.10\ \ \ Eqn\ 2$

From equation 1

$a = 133-c$

Putting this in equation 2

$9.50(133-c)+5.20c = 928.10\\1263.5-9.50c+5.20c=928.10\\1263.5-4.3c = 928.10\\-4.3c = 928.10-1263.5\\-4.3c=-335.4\\\frac{-4.3c}{-4.3} = \frac{-335.4}{-4.3}\\c = 78$

Putting c = 78 in equation 1

$a+78 = 133\\a = 133-78\\a = 55$

Hence,

55 adult tickets were sold on Friday.

4 0

2 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

2 years ago

Can someone please help me with this?

patriot [66]

The answer is 3y=-9x-6

7 0

2 years ago

Write an equation of the line in slope-intercept form. (Pls answer fast)<br><br> Y=

nydimaria [60]

Answer:

$y = - \frac{1}{4} x + 3$

6 0

2 years ago

Given the function: f(x) = 2x3 - 3/2x, find f(-4).

Sergeu [11.5K]

Answer:

D) -122

Step-by-step explanation:

Given:

$f(x) = 2x^3-\frac{3}2x$

We need to find f(-4).

Now by substituting the value of x =-4 in above equation we get;

$f(-4)= 2\times(-4)^3-\frac{3}{2}\times(-4)\\$

By using PEDMAS we will solve for exponential function first we get;

Now we know that $(-4)^3=-64$

So we get;

$f(-4)= 2\times(-64)-\frac{3}{2}\times(-4)\\$

Now using PEDMAS we will solve division.

$f(-4)= 2\times(-64)-1.5\times(-4)$

Now using PEDMAS we solve multiplication.

$f(-4)= -128+6$

Finally by subtraction we get;

$f(-4)= -122$

5 0

2 years ago