All categories

3 years ago

*Will give brainiest. Tina invests $1000 in an account that pays an interest rate of 6.75%. A=P(1+r)^t.

2 answers:

5 0

A=P(1+r)^t
18600=1000(1+0.0675)^t
18600/1000= (1.0675)^t
18.6= (1.0675)^t
log 18.6 = t log 1.0675
t= log 18.6/ log 1.0675
t=44.75 years

leonid [27]3 years ago

4 0

T=log(18.6)/log(1.0675)=???

You might be interested in

Type the correct answer in each box. use numerals instead of words. students in a reading class have three options for their boo

Alexus [3.1K]

The answers to the questions on how she would use the colored chips are given here:

brochure 4 blue chips

writing a new article 6 yellow chips

designing a computer presentation 10 red chips

<h3>How to solve for the values</h3>

For the first percentage

20 / 100 = x/20

= 20 * 20 = 100x

400 = 100x

divide through by 100

x = 400/100

x = 4

For the second percentage

30/100 = x/20

600 = 100x

x = 600/100

x = 6

For the last

50/100 = x/20

1000 = 100x

x = 10

The proof is that 6 + 4 + 10 still gives us 20

Read more on percentages here: brainly.com/question/24877689

#SPJ1

4 0

2 years ago

Find the value of x<br> (I need help with both of them pls)

Aleks04 [339]

Answer:

1. x = 21

2. x = 19

Step-by-step explanation:

Since the interior angles of all quadrilaterals add up to 360, all we have to do is add them up and set them equal to 360. Combine like terms and we get

7x + 213 = 360

Which further turns to 7x = 147

So x = 21

For the second problem, it's the same thing

By the end we get 2x + 8 = 46

2x = 38

x = 19

Hope this helps

6 0

2 years ago

An aeroplane X whose average speed is 50°km/hr leaves kano airport at 7.00am and travels for 2 hours on a bearing 050°. It then

Zigmanuir [339]

Answer:

(a)123 km/hr

(b)39 degrees

Step-by-step explanation:

Plane X with an average speed of 50km/hr travels for 2 hours from P (Kano Airport) to point Q in the diagram.

Distance = Speed X Time

Therefore: PQ =50km/hr X 2 hr =100 km

It moves from Point Q at 9.00 am and arrives at the airstrip A by 11.30am.

Distance, QA=50km/hr X 2.5 hr =125 km

Using alternate angles in the diagram:

$\angle Q=110^\circ$

(a)First, we calculate the distance traveled, PA by plane Y.

Using Cosine rule

$q^2=p^2+a^2-2pa\cos Q\\q^2=100^2+125^2-2(100)(125)\cos 110^\circ\\q^2=34175.50\\q=184.87$ km$

SInce aeroplane Y leaves kano airport at 10.00am and arrives at 11.30am

Time taken =1.5 hour

Therefore:

Average Speed of Y

$=184.87 \div 1.5\\=123.25$ km/hr\\\approx 123$ km/hr (correct to three significant figures)$

(b)Flight Direction of Y

Using Law of Sines

$\dfrac{p}{\sin P} =\dfrac{q}{\sin Q}\\\dfrac{125}{\sin P} =\dfrac{184.87}{\sin 110}\\123 \times \sin P=125 \times \sin 110\\\sin P=(125 \times \sin 110) \div 184.87\\P=\arcsin [(125 \times \sin 110) \div 184.87]\\P=39^\circ $ (to the nearest degree)$