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3 years ago

Find the area of the

Mathematics

1 answer:

Alex73 [517]3 years ago

5 0

162 m^2

you can make a triangle in the corner with legs length s/2 and the angles are 45 degrees. in a 45 45 90 triangle, the legs are h/sqrt(2) where h is the hypotenuse. this means that s/2=9/2 sqrt(2), s=9 sqrt(2)

and s^2=162

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If the simple interest on $3000 for 8 years is $1680, then what is the interest rate?

Fiesta28 [93]

R=i/pt
R=1680/3000*8=1,680÷24,000
=0.07*100=7%

3 0

3 years ago

Current Attempt in Progress

monitta

If at June 1, 2022, Sheffield Corp. had an Accounts Receivable balance of $ 18,900. During the month, the company had credit sales of $

24,500 and collected Accounts Receivable of $ 28,500. The balance in Accounts Receivable at June 30, 2022 will be: $14,900

Using this formula

2022 Balance in Accounts Receivable=Accounts Receivable balance as on June 1, 2022+ Credit sales -Collected Accounts Receivable

Let plug in the formula

2022 Balance in Accounts Receivable=$18,900+$24,500-$28,500

2022 Balance in Accounts Receivable=$14,900

Inconclusion if at June 1, 2022, Sheffield Corp. had an Accounts Receivable balance of $ 18,900. During the month, the company had credit sales of $

24,500 and collected Accounts Receivable of $ 28,500. The balance in Accounts Receivable at June 30, 2022 will be: $14,900

Learn more here:

brainly.com/question/24297448

3 0

3 years ago

What is the equation for a line that passes through the points (5,-4) and (-10,17

maria [59]

Answer:

The equation of the line is 7 x +5 y = 15.

Step-by-step explanation:

Here the given points are ( 5, -4) & ( -10, 17) -

Equation of a line whose points are given such that

( $x_{1},y_{1}$ ) & ( $x_{2}, y_{2}$ )-

y - $y_{1}$ = $\frac{y_{2} - y_{1} }{x_{2} - x_{1}}$ ( x - $x_{1}$ )

i.e. y - (-4)= $\frac{17 - (-4)}{-10 - 5}$ ( x- 5)

y + 4 = $\frac{17 + 4}{ -15}$ ( x - 5)

y + 4 = $\frac{21}{-15}$ ( x - 5 )

( y + 4) = $\frac{7}{- 5}$ ( x - 5)

5 (y + 4 ) = - 7 (x - 5 )

5 y + 20 = -7 x + 35

7 x + 5 y = 15

Hence the equation of the required line whose passes trough the points ( 5, -4) & ( -10, 17) is 7 x + 5 y = 15.

7 0

3 years ago

In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (En

mina [271]

Answer:

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

$\frac{dy}{dx} = 3t^{2} - 2t - 27$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$3t^{2} - 2t - 27 = 0$

$a = 3, b = -2, c = -27$

Then

$\bigtriangleup = b^{2} - 4ac = (-2)^{2} - 4*3*(-27) = 328$

$t_{1} = \frac{-(-2) + \sqrt{328}}{2*3} = 3.35$

$t_{2} = \frac{-(-2) - \sqrt{328}}{2*3} = -2.685$

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

$x = t^{3} - 3t = (3.35)^{3} - 3*3.35 = 27.55$

$y = t^{2} - 4 = (3.35)^2 - 4 = 7.22$

The first point is (27.55, 7.22)

t = -2.685

$x = t^{3} - 3t = (-2.685)^3 - 3*(-2.685) = -11.3$

$y = t^{2} - 4 = (-2.685)^2 - 4 = 3.21$

The second point is (-11.3, 3.21).

8 0

3 years ago

Find the absolute value of -6/3/5

sashaice [31]

(|-6| / 3) / 5 = 0.4

6 0

3 years ago