Ask question

Ask question

All categories

IgorLugansk [536]

3 years ago

10

LEIM deposits some money in an account which pays 5% simple interest per year, and twice that amount in an account paying 4% sim

ple interest per year. If he collects $533.00 in interest after one year, how much did he deposit in the account that has a 5% rate?

2 answers:

Tamiku [17]3 years ago

8 0

Answer:

266.5

Step-by-step explanation:

$533.00* 5%=266.5

olga nikolaevna [1]3 years ago

7 0

The answer is 266.5 hope this helps

You might be interested in

Some one cans help me to find the answer about the whole shape and tell me how do you get the answer step by step please I need

Naya [18.7K]

First let’s call the square shape A and the rectangle shape B.

To find the area of shape A you do 24 x 24 = 576cm squared

To find the area of shape B you do 40 x 24 = 960cm squared

To find the total area you just add the two areas together
576 + 960 = 1536cm squared

4 0

3 years ago

18 55 -101 196 55 18 22 X-22 57 88

choli [55]

Answer:

maybe a that's my answer

8 0

2 years ago

Choose the inverse of y=X2-2

pickupchik [31]

Answer:

26

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Which table represents y as. Function of x?

RoseWind [281]

I would say j but i can’t get my mind on which one

6 0

2 years ago

If sinθ = -1/2 and θ is in Quadrant III, then tanθ = _____.

Gnom [1K]

Answer: $\tan \theta=\dfrac{1}{\sqrt3}.$

Step-by-step explanation: Given that

$\sin\theta=-\dfrac{1}{2}$ and $\theta$ lies in Quadrant III.

We are to find the value of $\tan \theta.$

We will be using the following trigonometric identities:

$(i)~sin^2\theta+\cos^2\theta=1,\\\\(ii)~\dfrac{\sin\theta}{\cos{\theta}}=\tan \theta.$

We have

$\tan\theta\\\\\\=\dfrac{\sin\theta}{\cos\theta}\\\\\\=\dfrac{\sin\theta}{\pm\sqrt{1-\sin^2\theta}}\\\\\\=\pm\dfrac{-\frac{1}{2}}{\sqrt{1-\left(\frac{1}{2}\right)^2}}\\\\\\=\pm\dfrac{\frac{1}{2}}{\sqrt{1-\frac{1}{4}}}\\\\\\=\pm\dfrac{\frac{1}{2}}{\frac{\sqrt3}{2}}\\\\\\=\pm\dfrac{1}{\sqrt3}.$

Since $\theta$ lies in Quadrant III, so tangent will be positive.

Thus,

$\tan \theta=\dfrac{1}{\sqrt3}.$

8 0

3 years ago

Read 2 more answers