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2 years ago

A total of $5000 is invested: part at 7 % and the remainder at 15 %. How much is invested at each rate if

1 answer:

6 0

0.07 (5000-x) + 0.15x= 480
Solve for x
350-0.07x+0.15x=480
0.08x=130
X=1625 invested at 15%
(5000-1625)=3375 invested at 7%

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Will give 30 points!<br> What is a expanded notation?

gavmur [86]

Answer:

It is the number written out to show each digit

Step-by-step explanation:

Let's say you have the number 459

you write out the hundredths place, which is 4 times 100

you write out the tenths place which is 5 times 10

and you write out the oneths place which is 9 times 1

the full thing written out is 459 = 100×4 + 5×10 + 9 × 1

4 0

3 years ago

George and Paula are running around a circular track. George starts at the westernmost point of the track, and Paula starts at t

arsen [322]

Answer:

George is 43.20 ft East of his starting point.

Step-by-step explanation:

Let Paula's speed be x ft/s

George's speed = 9 ft/s

Note that speed = (distance)/(time)

Distance = (speed) × (time)

George takes 50 s to run a lap of the track at a speed of y ft/s

Meaning that the length of the circular track = y × 50 = 50y ft

George and Paula meet 14 seconds after the start of the run.

Distance covered by George in 14 seconds = 9 × 14 = 126 ft

Distance covered by Paula in 14 seconds = y × 14 = 14y ft

But the sum of the distance covered by both runners in the 14 s before they first meet each other is equal to the length of the circular track

That is,

126 + 14y = 50y

50y - 14y = 126

36y = 126

y = (126/36) = 3.5 ft/s.

Hence, Paula's speed = 3.5 ft/s

Length of the circular track = 50y = 50 × 3.5 = 175 ft

So, in 4 minutes (240 s), with George running at 9 ft/s, he would have ran a total distance of

9 × 240 = 2160 ft.

2160 ft around a circular track of length 175 ft, means that George would have ran a total number of laps (2160/175) = 12.343 laps.

Breaking this into 12 laps and 0.343 of a lap from the starting point. 0.343 of a lap = 0.343 × 175 = 60 ft

So, 60 ft along a circular track subtends an angle θ at the centre of the circle.

Length of an arc = (θ/360°) × 2πr

2πr = total length of the circular track = 175

r = (175/2π) = 27.85 ft

Length of an arc = (θ/360) × 2πr

60 = (θ/360°) × 175

(θ/360°) = (60/175) = 0.343

θ = 0.343 × 360° = 123.45°

The image of this incomplete lap is shown in the attached image,

The distance of George from his starting point along the centre of the circular track = (r + a)

But, a can be obtained using trigonometric relations.

Cos 56.55° = (a/r) = (a/27.85)

a = 27.85 cos 56.55° = 15.35 ft

r + a = 27.85 + 15.35 = 43.20 ft.

Hence, George is 43.20 ft East of his starting point.

Hope this Helps!!!

6 0

3 years ago

In a triangle ABC, measure of angle B is 90 degrees. AB is 3x-2 units and BC is x+3. If the area of the triangle is 17 sq cm, fo

Scorpion4ik [409]

Answer:

$x=\frac{8}{3}\ cm$

Step-by-step explanation:

we know that

The area of the right triangle ABC is equal to

$A=\frac{1}{2}(AB)(BC)$

we have

$A=17\ cm^2$

$AB=(3x-2)\ cm$

$BC=(x+3)\ cm$

substitute the values

$17=\frac{1}{2}(3x-2)(x+3)$

$34=(3x-2)(x+3)$

$34=3x^2+9x-2x-6$

$3x^2+7x-6-34=0$

$3x^2+7x-40=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$3x^2+7x-40=0$

$a=3\\b=7\\c=-40$

substitute in the formula

$x=\frac{-7(+/-)\sqrt{7^{2}-4(3)(-40)}} {2(3)}$

$x=\frac{-7(+/-)\sqrt{529}} {6}$

$x=\frac{-7(+/-)23} {6}$

$x=\frac{-7(+)23} {6}=\frac{16}{6}=\frac{8}{3}$

$x=\frac{-7(-)23} {6}=-5$

therefore

The solution is

$x=\frac{8}{3}\ cm$

6 0

3 years ago

Could someone help me find the slope?

shutvik [7]

2 over 3 you're welcome :)

3 0

3 years ago

A triangle has side lengths of 24 in., 32 in., and 40 in. Is the triangle acute, obtuse, or right? A) right. B) obtuse. C) acute

Feliz [49]

24, 32 , 40. Divide through by 8

3, 4, 5,

The 3 , 4, 5 reminds of the Pythagorean triple, which is a right angled triangle.

So a triangle with sides 24, 32, 40 is a right angled triangle.

Option A.

8 0

3 years ago

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