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e-lub [12.9K]
3 years ago
13

Kunio puts $3,300.00 into his savings bonds that pay a simple interest rate of 2.5%. How much money will the bonds be worth at t

he end of 5 years?
Mathematics
1 answer:
vagabundo [1.1K]3 years ago
7 0

Answer:

3,712.50

Step-by-step explanation:

2.5% divided by 100 = .025 which is the percent in decimal form.

Use the Equation Principal = (Amount Invested * Rate * Time)


3,300.00 * 0.025 * 5 = 412.50 earned

3,300 + 412.50 = 3,712.50 total

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So how much does she spend in all right
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Help!!! please what's is the figure​
S_A_V [24]

Hello from MrBillDoesMath!

Answer:

108 cm^2

Discussion:

The area of the logo  =

area of Trapezoid on top (I)  + area of triangle on bottom (II)

I  = (1/2) h ( b1 + b2)            h = altitude, b1, b2 = bases

  = (1/2)  6  ( 6  + 12)

  = (1/2) 6 ( 18)

  = 54

II  = (1/2) b h

   = (1/2) (12) ( 15 - 6)               => altitude of triangle = 15 - 6 NOT (1/2) 15

   =  54

I + II= 54 + 54 = 108

Thank you,

MrB

3 0
3 years ago
Read 2 more answers
Nick planted a cedar tree. It grew 10 cm in ½ of a year. What will the height of the tree be at the end of 6 years?
user100 [1]

Answer:

120 cm

Step-by-step explanation:

1/2 is 6 months and 6 years are 72 months.

Using the rule of three, 10 cm is to 6 months what x is for 72 months so

x = (72 * 10) / 6 = 120

7 0
3 years ago
Please solve the challenge problem J
Verdich [7]

Answer: (7x^2+144)/x^2

Step-by-step explanation:

f(h(x))=

f(12/x)=

(12/x)^2+7=

12^2/x^2 + 7=

144/x^2 +7=

144/x^2 +7*x^2/x^2=

144/x^2+7x^2/x^2=(7x^2+144)/x^2

6 0
2 years ago
The equation gives the speed at impact, V metres per second, of an object dropped from a height of h metres. SHOW WORK From what
devlian [24]

Answer:

h = 17.65 m

Step-by-step explanation:

The given equation gives the speed at impact :

v=\sqrt{2gh}

h is height form where the object is dropped

Put v = 18.6 m/s in the above equation.

h=\dfrac{v^2}{2g}\\\\h=\dfrac{(18.6)^2}{2\times 9.8}\\\\h=17.65\ m

So, the object must be dropped from a height of 17.65 m.

4 0
3 years ago
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