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3 years ago

The price of jeans was reduced $6 per week for 7 weeks. By how much did the price of the jeans change over 7 weeks?

1 answer:

5 0

Just multiply 6 by 7 since the price is reduced $6 per week.

$6.00 × 7 = $42.00

Your final answer is $42

I hope this helped!!

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Using the slope formula, find the slope of the line through the given points.

UNO [17]

Answer:

Step-by-step explanation:

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

= (-5--5)/ (9--8)

= (-5+5)/(9+8)

= 0/17

The slope it 0

This is a horizontal line because the y value doesn't change. If the y value doesn't change, the slope it zero.

4 0

3 years ago

Hello answer and I will give you brainliest:No cheating!

stich3 [128]

The answer is 58,875

6 0

2 years ago

Best decimal equivalent to 5/8

Igoryamba

The best decimal equivalent is 10/16

6 0

3 years ago

Read 2 more answers

An object at the top of a building with height 110 feet is thrown upward with an initial speed of 23 ft/s. Find its position z(t

alina1380 [7]

Answer:

Step-by-step explanation:

Given

Height of building is $h=110\ ft$

Initial upward velocity $u=23\ ft/s$

height of object is given by

$z(t)=ut+\frac{1}{2}gt^2$

but building is already at a height of 110 ft so z(t) must be given by

$Z(t)=ut+\frac{1}{2}(-g)t^2+110$

$Z(t)=23\times t-\frac{1}{2}\times 9.8\times t^2$

$Z(t)=-16t^2+23t+110$

Time when it hits is given by when Z(t)=0[/tex]

$-16t^2+23t+110=0$

$t=\frac{-23\pm \sqrt{(23)^2-4\times (-16)\times 110}}{2\times (-16)}$

$t=\frac{-23\pm 87}{-32}$

$t=3.43\ s$

6 0

3 years ago

A chemist wants to make 100 liters of a 44% acid solution. She has solutions that are 20% acid and 60% acid.

alexgriva [62]

the first solution is 20% acid, and say we'll be using "x" liters, so how many liters of just acid are in it? well 20% of "x" or namely 0.2x. Likewise for the 60% acid solution, if we had "y" liters of it, the amount of only acid in it is 0.6y.

$\begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ \textit{1st solution}&x&0.20&0.2x\\ \textit{2nd solution}&y&0.60&0.6y\\ \cline{2-4}&\\ mixture&100&0.44&44 \end{array}~\hfill \begin{cases} x+y=100\\\\ 0.2x+0.6y=44 \end{cases}$

$x+y=100\implies y=100-x~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.2x+0.6(100-x)=44} \\\\\\ 0.2x+60-0.6x=44\implies -0.4x+60=44\implies -0.4x=-16 \\\\\\ x=\cfrac{-16}{-0.4}\implies \boxed{x=40}~\hfill \boxed{\stackrel{100-40}{y=60}}$

7 0

2 years ago