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AlexFokin [52]
3 years ago
8

PLEASE HELP Joy invest a total of 6,500 in two accounts The first account earned a rate of return of 13% after year. However, th

e second account suffered a 10% loss in the same time period. At the end of one year, the total amount of money gained was -$75 how much was invested into each account

Mathematics
1 answer:
Westkost [7]3 years ago
8 0

Answer:

First account= 2500

Second account= 4000

Step-by-step explanation:

Total amount invested: 6500

Amount in first account: x

Amount in second account: y

x+y=6500

1.13x+0.9y=6500-75=6425

You have two equation now just solve the system of equation through either substitution or elimination.

Then you'll get the answer

x= 2500

y= 4000

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Simplify,write equivalent expressions. 1/6(15a+20b)
lutik1710 [3]
1/6* (15a+20b)
= 1/6* 15a + 1/6* 20b (Distributive property)
= 2.5a + 10/3*b

The final answer is 2.5a+ 10/3b

Hope this helps~
6 0
3 years ago
The area of a triangle is 20 cm², and the
grandymaker [24]

Answer:

Heavenly,  we need the forumula for the area of a triangle, do you know it?

Step-by-step explanation:

Area of a triangle is (1/2 ) * B*H

where B = base

H = height

then

A = 1/2* B * H

A is given to be 20

B = b

H = b+4

plug all those into the formula,

20 = (1/2)*b*(b+4)

use your algebra skillz  :P

20 = 1/2 *( b^{2} + 4b)

20*2 = b^{2} + 4b

40 = b^{2} + 4b

0 = b^{2} + 4b-40

rewrite the above so it's in x format

x^{2} + 4x -40 = 0

use the quadratic formula to find  x  , and for our problem, remember that x represents the base,  or b,  but don't mix it up with the b in the quadatic fomula,  with the b of the triangle, which is the base,  they are differnt "b"s   :P

X = (b±\sqrt{b^{2}-4*a*c } ) / 2*a

a = 1

b = 4

c = -40

x = (4±\sqrt{4^{2}-4*1*(-40) }  ) /2*1

x = ( 4±\sqrt{16+160} ) / 2

x = (4±\sqrt{176} ) / 2

x = (4±13.266499)/2

x= 2±6.633249

x = 2 + 6.633249 = 8.633

b= 8.633 cm

5 0
2 years ago
How do you simplify 16^2/24^7
quester [9]

Answer:

1/(2¹³·3⁷)

Step-by-step explanation:

                                                16²/24⁷

Factor the terms                  = (2⁴)²/(8×3)⁷

Apply the Power Rule        = 2⁸/(2³×3)⁷

Apply the Power Rule        = 2⁸/(2²¹ × 3⁷)

Apply the Quotient Rule    = 1/(2¹³·3⁷)

5 0
3 years ago
Polygon F has an area of 36 square units. Aimar drew a scaled version of Polygon F and labeled it Polygon G. Polygon G has an ar
Free_Kalibri [48]

Answer:

1/3

Step-by-step explanation:

The area of Polygon GGG is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF.

Each side of Polygon FFF was multiplied by a certain value, known as the scale factor , to result in an area that is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF.

[Show me an example of how scale factor affects area]

\dfrac1{10}  

start fraction, 1, divided by, 10, end fraction

 

 

\begin{aligned} A &= \left(l\times\dfrac1{10}\right)\times\left(w\times\dfrac1{10}\right) \\ \\ A&= l\times w\times\dfrac1{10}\times\dfrac1{10} \\ \\ A&= lw \times \left(\dfrac1{10}\right)^2\end{aligned}  

 

 

 

 

 

 

 

 

\dfrac1{10}  

start fraction, 1, divided by, 10, end fraction\left(\dfrac1{10}\right)^2  

 

left parenthesis, start fraction, 1, divided by, 10, end fraction, right parenthesis, start superscript, 2, end superscript

Hint #22 / 3

The area of a polygon created with a scale factor of \dfrac1x  

x

1

​  start fraction, 1, divided by, x, end fraction has \left(\dfrac1{x}\right)^2(  

x

1

​  )  

2

left parenthesis, start fraction, 1, divided by, x, end fraction, right parenthesis, start superscript, 2, end superscript the area of the original polygon:

\left(\text{scale factor}\right)^2=\text{fraction of the area the scale copy has}(scale factor)  

2

=fraction of the area the scale copy hasleft parenthesis, s, c, a, l, e, space, f, a, c, t, o, r, right parenthesis, start superscript, 2, end superscript, equals, f, r, a, c, t, i, o, n, space, o, f, space, t, h, e, space, a, r, e, a, space, t, h, e, space, s, c, a, l, e, space, c, o, p, y, space, h, a, s

The area of Polygon GGG is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF. Let's substitute \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction into the equation to find the scale factor.

\left(\dfrac1{?}\right)^2=\dfrac19(  

?

1

​  )  

2

=  

9

1

​  left parenthesis, start fraction, 1, divided by, question mark, end fraction, right parenthesis, start superscript, 2, end superscript, equals, start fraction, 1, divided by, 9, end fraction

The scale factor is \dfrac13  

3

1

​  start fraction, 1, divided by, 3, end fraction.

Hint #33 / 3

Aimar used a scale factor of \dfrac13  

3

1

​  start fraction, 1, divided by, 3, end fraction to go from Polygon FFF to Polygon GGG.

4 0
3 years ago
Read 2 more answers
Honors segment one activity part A
Delicious77 [7]

Answer:

I couldn't answer on your most recent question so here :) Good luck

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