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Reika [66]
2 years ago
11

A bakery’s balance sheet shows the following assets (in thousands): cash, $356; accounts receivable, $600; inventory, $510; prop

erty, $1206; other assets, $535. The liabilities were: loans and accounts payable, $600; other current liabilities, $225; long-term debt, $180. Find the owners’ equity.
$2,202,000

$1,005,000

$3,207,000

$4,213,000
Mathematics
1 answer:
OLga [1]2 years ago
5 0
The owners' equity will be 1,005,000
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Which is a solution to the equation? (x −2)(x + 5) = 18 x = −10 x = −7 x = −4 x = −2
Leona [35]

Answer:

Solutions:

x=-7

x=4

Step-by-step explanation:

(x-2)(x + 5) = 18

Expanding the factored form:

x^2+5x-2x-10=18

x^2+3x-28=0

Factoring the second-degree polynomial:

(x+7)(x-4)=0

This is true for

x=-7 \text{ or } x=4

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2 years ago
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6 is yes because they have the same angles and lengths while 7 is no because the angles are not the same
8 0
2 years ago
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
2 years ago
The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines si
natta225 [31]

Answer:

a) y = 7.74*x + 7.5

b)  y = 1.148*x + 6.036

Step-by-step explanation:

Given:

                                  f(x) = 6 - 10*x^2

                                  g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

                                f'(x) = -20*x

                                g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

                                g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

                                m = (g(x_o) - f(x)) / (x_o - x)

                                m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

                                m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

                                m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

                                m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

                                m = f'(x) = g'(x_o)

- We will develop the first expression:

                                m = f'(x)

                                ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1.                          (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

                              m = g'(x_o)

                              ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

                              -2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2                       -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

                              -20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

                              -22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors:       20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

                              20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

                               (x - x_o)(20*x - 2*x_o + 4) = 0  

                               x = x_o   ,     x_o = 10x + 2    

- For x_o = 10x + 2  ,

                               (g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

                                (8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

                                (-90*x^2 + 2) = -180*x^2 - 40*x

                                90*x^2 + 40*x + 2 = 0  

- Solve the quadratic equation above:

                                 x = -0.0574, -0.387      

- Largest slope is at x = -0.387 where equation of line is:

                                  y - 4.502 = -20*(-0.387)*(x + 0.387)

                                  y = 7.74*x + 7.5          

- Other tangent line:

                                  y - 5.97 = 1.148*(x + 0.0574)

                                  y = 1.148*x + 6.036

6 0
2 years ago
What is the equation of this circle in standard form?
Musya8 [376]

Answer:

D

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Here (h, k ) = (2, 3 ) and r = 6, thus

(x - 2)² + (y - 3)² = 6², that is

(x - 2)² + (y - 3)² = 36 → D

3 0
2 years ago
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