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3 years ago

5

Dave is painting a wall in his living room a bright red color. so far, he has painted 1/5 of the wall and has used 1/2 gallon of

paint. how many gallons of paint does he need to cover the entire wall

1 answer:

uysha [10]3 years ago

6 0

2 and 1/2 gallons of paint because if 2/5 is painted with 1 gallon, then 4/5 would be 2 gallons and the last 1/5 would be half a gallon.

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16/25+8/5+4+10…+a10<br> what’s the sum of the series

sweet-ann [11.9K]

Answer:

correct answer is 406/25+a10

Step-by-step explanation:

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2 years ago

A box contains 14 books & 7 pieces of dusters. Each piece of duster weighs 0.2 pounds each book weighs 1.1 pounds. The box &

Olin [163]

Answer:

1.45

Step-by-step explanation:

weight of all books= 14×1.1= 15.4

weight of all duster=7×0.2= 1.4

weight of books and duster= 15.4+1.4=16.8

weight of the empty box = 18.25-16.8=1.45

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2 years ago

Can someone help me on this math problem ?

faltersainse [42]

Answer:

4/3

Step-by-step explanation:

Given two points, we can find the slope by using

m = (y2-y1)/(x2-x1)

= (20 - -16)/(7 - -20)

= (20+16)/(7+20)

= 36/27

=4/3

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3 years ago

Write the equation of the line in slope intercept from given the slope of the line is m=3 and the line passes through the point

inna [77]

Answer:

math way

Step-by-step explanation:

helps

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3 years ago

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The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.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 = <u> -1 000.00 </u>

P = 22 944.00

1 Jan 2016 A = £23 517.60

5 0

3 years ago