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aniked [119]
3 years ago
5

Use simple interest to find the ending balance​

Mathematics
1 answer:
Elena-2011 [213]3 years ago
6 0
Answer: 7788.50

Solving steps:
P = 7400, R = 10.5%, T = 6

F = P + P rt
P = 7400
R = 10.5%
T = 6

F = 7400 + 7400rt

7400 x (12 + 6 x 0.105/12

7400 x (12 + 0.63)/12

7400 x 12.63/12

93462/12

15577/12

15577/2 0.01
7788.50
Answer: 7788.50
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Answer:

9)    a = ¾, <u>vertex</u>: (-4, 2),  <u>Equation</u>: y = ¾|x + 4| + 2

10)  a = ¼, <u>vertex</u>: (0, -3),  <u>Equation</u>: y = ¼|x - 0| - 3

11)   a = -4,  <u>vertex</u>: (3,  1),   <u>Equation</u>: y = -4|x - 3| + 1

12)  a = 1,    <u>vertex</u>: (-2, -2),  <u>Equation</u>: y = |x + 2| - 2

Step-by-step explanation:

<h3><u>Note:</u></h3>

I could <u><em>only</em></u> work on questions 9, 10, 11, 12 in accordance with Brainly's rules. Nevertheless, the techniques demonstrated in this post applies to all of the given problems in your worksheet.

<h2><u>Definitions:</u></h2>

The given set of graphs are examples of absolute value functions. The <u>general form</u> of absolute value functions is: y = a|x – h| + k, where:

|a|  = determines the vertical stretch or compression factor (wideness or narrowness of the graph).

(h, k) = vertex of the function

x = h represents the axis of symmetry.

<h2><u>Solutions:</u></h2><h3>Question 9)  ⇒ Vertex: (-4, 2)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (0, 5) and substitute into the general form (equation):

y = a|x – h| + k

5 = a| 0 - (-4)| + 2

5 = a| 0 + 4 | + 2

5 = a|4| + 2

5 = 4a + 2

Subtract 2 from both sides:

5 - 2 = 4a + 2 - 2

3 = 4a

Divide both sides by 4 to solve for <em>a</em>:

\LARGE\mathsf{\frac{3}{4}\:=\:\frac{4a}{4}}

a = ¾

Therefore, given the value of a = ¾, and the vertex, (-4, 2), then the equation of the absolute value function is:

<u>Equation</u>:  y = ¾|x + 4| + 2

<h3>Question 10)  ⇒ Vertex: (0, -3)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (4, -2) and substitute into the general form (equation):

y = a|x – h| + k

-2 = a|4 - 0| -3

-2 = a|4| - 3

-2 = 4a - 3

Add 3 to both sides:

-2 + 3 = 4a - 3 + 3

1 = 4a  

Divide both sides by 4 to solve for <em>a</em>:

\LARGE\mathsf{\frac{1}{4}\:=\:\frac{4a}{4}}

a = ¼

Therefore, given the value of a = ¼, and the vertex, (0, -3), then the equation of the absolute value function is:

<u>Equation</u>:  y = ¼|x - 0| - 3

<h3>Question 11)  ⇒ Vertex: (3, 1)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (4, -3) and substitute into the general form (equation):

y = a|x – h| + k

-3 = a|4 - 3| + 1

-3 = a|1| + 1

-3 = a + 1

Subtract 1 from both sides to isolate <em>a</em>:

-3 - 1 = a + 1 - 1

a = -4

Therefore, given the value of a = -4, and the vertex, (3, 1), then the equation of the absolute value function is:

<u>Equation</u>:  y = -4|x - 3| + 1

<h3>Question 12)  ⇒ Vertex: (-2, -2)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (-4, 0) and substitute into the general form (equation):

y = a|x – h| + k

0 = a|-4 - (-2)| - 2

0 = a|-4 + 2| - 2

0 = a|-2| - 2

0 = 2a - 2

Add 2 to both sides:

0 + 2  = 2a - 2 + 2

2 = 2a

Divide both sides by 2 to solve for <em>a</em>:

\LARGE\mathsf{\frac{2}{2}\:=\:\frac{2a}{2}}

a = 1

Therefore, given the value of a = -1, and the vertex, (-2, -2), then the equation of the absolute value function is:

<u>Equation</u>:  y = |x + 2| - 2

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2 years ago
What is the area of the rectangle with the given dimensions?
rewona [7]

Answer:

F

Step-by-step explanation:

The area is equal to length times width.

Use the foil method to multiply (2x-5)(3x+3).

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Which table represents a proportional relationship?
Lena [83]

9514 1404 393

Answer:

  C)

Step-by-step explanation:

The table that has a constant ratio between y and x values is the one that represents a proportional relationship.

A) 2/4 ≠ 4/16

B) 1/1 ≠ 4/16

C) 6/8 = 12/16 = 18/24 = 30/40, a proportional relationship

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