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3 years ago

Paddy and Anna each invest $2000 for 5 years.

Mathematics

1 answer:

Serga [27]3 years ago

6 0

Answer: r= 1.22

Step-by-step explanation:

Formula for amount with simple interest = $P(1+rt)$

, where

P= principal value , r= rate of interest , t = time.

Given: P= $2000, t= 5 years, r= 1.25% = 0.0125

$A=2000(1+0.0125\times5)\\\\=2000(1.0625)=2125$

Formula to compute compound amount : $P(1+r)^t$

$=2000(1+r)^5$

When both have same worth then

$2000(1+r)^5=2125\\\\\\ (1+r)^5=\dfrac{2125}{2000}\\\\\\ (1+r)^5=1.0625$

taking log on both sides , we get

$5\ln (1+r)=\ln 1.0625\\\\\\ 5\ln (1+r)=0.0606246\\\\\\ \ln (1+r)=0.012125\\\\\\ 1+r=e^{0.005266}\\\\\\ 1+r=1.0122\\\\ r=0.0122\\\\ \\ r=1.22\%$

Hence, Value of r= 1.22

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By what percent will a fraction CHANGE(not increase!!) if it’s numerator is decreased by 60% and it’s denominator is decreased b

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2 years ago

Henry and Bailey plan to make key chains. They need foot of string and 5 beads for each key chain. Henry has $8 to buy string, a

Alona [7]

Answer:

72 Keychains

Step-by-step explanation:

120 feet = $3.25

divide 120 by 3.25 to find how much string can be bought with $1. The result is 36.923. Then multiply that number by the number of dollars allotted to buying string. 8*36.923= 295.384. This is the total number of feet of string they have. They use 1 foot for each key chain so they can make 295 keychains based on how much string they have.

35 beads = $2.59

Divide 35 by 2.59 to find how many beads can be bought with $1. The result is 13.514. Then multiply that number by the number of dollars allotted to buying beads. This comes out to 364.878. This means She can buy 364 beads. Each Key chain needs 5 beads so divide 364/5 getting a total of 72 keychains.

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5 0

2 years ago

Given two points P(sinθ+2, tanθ-2) and Q(4sin²θ+4sinθcosθ+2acosθ, 3sinθ-2cosθ+a). Find constant "a" and the corresponding value

vodomira [7]

Answer:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1, \frac{\sqrt{3}}{2} - 1$

Step-by-step explanation:

we are given two coincident points

$\displaystyle P( \sin(θ)+2, \tan(θ)-2) \: \text{and } \\ \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

since they are coincident points

$\rm \displaystyle P( \sin(θ)+2, \tan(θ)-2) = \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ )\cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

By order pair we obtain:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ) = \sin( \theta) + 2 \\ \\ \displaystyle 3 \sin( \theta) - 2 \cos( \theta) + a = \tan( \theta) - 2\end{cases}$

now we end up with a simultaneous equation as we have two variables

to figure out the simultaneous equation we can consider using substitution method

to do so, make a the subject of the equation.therefore from the second equation we acquire:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sinθ \cos(θ)+2a \cos(θ )= \sin( \theta) + 2 \\ \\ \boxed{\displaystyle a = \tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) } \end{cases}$

now substitute:

$\rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2 \cos(θ) \{\tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) \}= \sin( \theta) + 2$

distribute:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ)+4 \sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) - 6 \sin( \theta) \cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

collect like terms:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

rearrange:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) + 4 \cos ^{2} ( \theta) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + = \sin( \theta) + 2$

by Pythagorean theorem we obtain:

$\rm\displaystyle \displaystyle 4 - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) + 2$

cancel 4 from both sides:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) - 2$

move right hand side expression to left hand side and change its sign:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+\sin(θ ) - 4\cos( \theta) + 2 = 0$

factor out sin:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) - 4\cos( \theta) + 2 = 0$

factor out 2:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) + 2(- 2\cos( \theta) + 1 ) = 0$

group:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(- 2 \cos(θ)+1) = 0$

factor out -1:

$\rm\displaystyle \displaystyle - ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

divide both sides by -1:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

by Zero product property we acquire:

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) + 2 = 0 \\ \displaystyle2 \cos(θ) - 1= 0 \end{cases}$

cancel 2 from the first equation and add 1 to the second equation since -1≤sinθ≤1 the first equation is false for any value of theta

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) \neq - 2 \\ \displaystyle2 \cos(θ) = 1\end{cases}$

divide both sides by 2:

$\rm\displaystyle \displaystyle \displaystyle \cos(θ) = \frac{1}{2}$

by unit circle we get:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

so when θ is 60° a is:

$\rm \displaystyle a = \tan( {60}^{ \circ} ) - 2 - 3 \sin( {60}^{ \circ} ) + 2 \cos( {60}^{ \circ} )$

recall unit circle:

$\rm \displaystyle a = \sqrt{3} - 2 - \frac{ 3\sqrt{3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1$

when θ is 300°

$\rm \displaystyle a = \tan( {300}^{ \circ} ) - 2 - 3 \sin( {300}^{ \circ} ) + 2 \cos( {300}^{ \circ} )$

remember unit circle:

$\rm \displaystyle a = - \sqrt{3} - 2 + \frac{3\sqrt{ 3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = \frac{ \sqrt{3} }{2} - 1$

and we are done!

disclaimer: also refer the attachment I did it first before answering the question

5 0

2 years ago

Can someone please give me a detailed answer please.

lubasha [3.4K]

The area of a triangle ( one of the sides) is 1/2 x base x height
base = 16, height = 19

Area = 1/2 x 16 x 19 = 152 square units.

There are 4 sides: 152 * 4 = 608 square units
The area of the bottom square is 16 x 16 = 256 square units.

Total area = 608 + 256 = 864 square units.

The answer is D.

5 0

3 years ago

PLEASE HELP ME I NEED THE PRIME FACTORIZATION!!!!!!!!!!!!!!!

docker41 [41]

Hey there :)

To find the Greatest Common Factor ( GCF ) of 36, 40 and 45, let us first separate:
Factors of 36: 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
Factors of 40: 1 , 2 , 4 , 5 , 8 , 10 , 20 , 40
Factors of 45: 1 , 3 , 5 , 9 , 15 , 45

Therefore, the GCF is 1 since only that is common in all three.

To find the Least Common Multiple of 36, 40 and 45, we see which multiplication matches in the least:
36 × 1 = 36 | 40 × 1 = 40 | 45 × 1 = 45
36 × 2 = 72 | 40 × 2 = 80 | 45 × 2 = 90
36 × 3 = 108 | 40 × 3 = 120 | 45 × 3 = 135
36 × 4 = 144 | 40 × 4 = 160 | 45 × 4 = 180
36 × 5 = 180 | 40 × 5 = 200 | 45 × 5 = 225
36 × 6 = 216 | 40 × 6 = 240 | 45 × 6 = 270
36 × 7 = 252 | 40 × 7 = 280 | 45 × 7 = 315
36 × 8 = 288 | 40 × 8 = 320 | 45 × 8 = 360
36 × 9 = 324 | 40 × 9 = 360
36 × 10 = 360

Therefore, the LCM is 360

4 0

3 years ago