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3 years ago

WILL GIVE BRAINLIEST Which of the following are true about the given rational equations? CHECK ALL THAT APPLY!

Mathematics

1 answer:

Juliette [100K]3 years ago

6 0

Answer:

the first one is true:)

Step-by-step explanation:

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a company declared a dividend of $1.50 per share. john has 600 shares in the company. how much will she get?

Harman [31]

Answer:

$900

Step-by-step explanation:

So if each individual share is $1.50 we can presume that we need to multiply both factors. 1 x 600 is 600 and .50 x 600 is 300. 600 + 300 = 900. John will get $900 in shares profit.

3 0

2 years ago

Please help ASAP this is 6th grade math plz

lianna [129]

Answer:

3 ft times x ft

Step-by-step explanation:

You don't know what the x is so you can keep it as a x.You are going to multiply it so it is 3 ft times x ft

6 0

3 years ago

30 points!

Andrei [34K]

Its the first one. Beacuse she gains speed then slows down and stops but then starts up again then slows down. Hope this helps!

7 0

3 years ago

Read 2 more answers

Solve the inequality x - 3 > 7

Katen [24]

Answer:

x > 10

Step-by-step explanation:

Given

x - 3 > 7 ( add 3 to both sides )

x > 10

4 0

2 years ago

Read 2 more answers

On the 1st January 2014 Carol invested some money in a bank account.

Ghella [55]

Answer:

$\large \boxed{\text{\pounds 23 360.00}}$

Step-by-step explanation:

The formula for the accrued amount from compound interest is

$A = P \left(1 + \dfrac{r}{n}\right)^{nt}$

1. Amount in account on 1 Jan 2015

(a) Data:

a = £23 517.60

r = 2.5 %

n = 1

t = 1 yr

(b) Calculations:

r = 0.025

$\begin{array}{rcl}23517.60 & = & P\left (1 + \dfrac{r}{n}\right)^{nt}\\\\& = & P\left (1 + \dfrac{0.025}{1}\right)^{1\times1}\\\\& = & P (1 + 0.025)\\ & = & 1.025 P\\P & = & \dfrac{23517.60 }{1.025} \\\\& = & 22 944.00 \\\end{array}$

The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.

She must have had £23 944 in her account on 1 Jan 2015.

(2) Amount originally invested

(a) Data

A = £23 944.00

$\begin{array}{rcl}23 944.00 & = & 1.025 P\\P & = & \dfrac{23 944.00 }{1.025} \\\\& = & \mathbf{23 360.00} \\\end{array}\\\text{Carol originally invested $\large \boxed{\textbf{\pounds23 360.00}}$ in her account.}$

3. Summary

1 Jan 2014 P = £23 360.00

1 Jan 2015 A = 23 944.00

Withdrawal = -1 000.00

P = 22 944.00

1 Jan 2016 A = £23 517.60

5 0

2 years ago