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3 years ago

10

Please help me fast thank you very much.

2 answers:

Anika [276]3 years ago

7 0

Answer:

-6

Step-by-step explanation:

ankoles [38]3 years ago

7 0

Answer:

-6

Step-by-step explanation:

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1.) Your 3 year investment of $20,000 received 5.2% interested compounded semi annually. What is your total return? ASW

Stella [2.4K]

Let's begin by listing out the information given to us:

Principal (p) = $20,000

Interest rate (r) = 5.2% = 0.052

Number of compounding (n) = 2 (semi annually)

Time (t) = 3 years

The total return is calculated as shown below:

A = p(1 + r/n)^nt

A = 20000(1 + 0.052/2)^2*3 = 20000(1 + 0.026)^6

A = 20000(1.1665) = 23,330

A = $23,330

4 0

1 year ago

Please help me quickly

hichkok12 [17]

Answer:

Convenience sampling

Step-by-step explanation:

5 0

3 years ago

Roger can type 1530 words in 1/2 hour. How many words can he type per<br> minute?<br> PLS HELP ME

Lera25 [3.4K]

Answer:

25.5

Step-by-step explanation:

1530 divided by 60 is 25.5 and 25.5 multiplied by 60 is 1530.

6 0

2 years ago

The coordinates G(7,3), H(9, 0), (5, -1) form what type of polygon?

Marrrta [24]

Answer:

Is an acute triangle

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

G(7,3), H(9, 0), I(5, -1)

step 1

Find the distance GH

substitute in the formula

$d=\sqrt{(0-3)^{2}+(9-7)^{2}}$

$d=\sqrt{(-3)^{2}+(2)^{2}}$

$GH=\sqrt{13}\ units$

step 2

Find the distance IH

substitute in the formula

$d=\sqrt{(0+1)^{2}+(9-5)^{2}}$

$d=\sqrt{(1)^{2}+(4)^{2}}$

$IH=\sqrt{17}\ units$

step 3

Find the distance GI

substitute in the formula

$d=\sqrt{(-1-3)^{2}+(5-7)^{2}}$

$d=\sqrt{(-4)^{2}+(-2)^{2}}$

$GI=\sqrt{20}\ units$

step 4

Verify what type of triangle is the polygon

we know that

If applying the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$ ----> is a right triangle

$c^{2}> a^{2}+b^{2}$ ----> is an obtuse triangle

$c^{2}< a^{2}+b^{2}$ ----> is an acute triangle

where

c is the greater side

we have

$c=\sqrt{20}\ units$

$a=\sqrt{17}\ units$

$b=\sqrt{13}\ units$

substitute

$c^{2}= (\sqrt{20})^{2}=20$

$a^{2}+b^{2}=(\sqrt{17})^{2}+(\sqrt{13})^{2}=30$

therefore

$c^{2}< a^{2}+b^{2}$

Is an acute triangle

6 0

3 years ago

Read 2 more answers

Square root of 141 '

Aleksandr [31]

Approximately 11.87434

8 0

3 years ago