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Gnesinka [82]
3 years ago
8

Darren is trying to improve his typing speed. On his first practice assignment, he types 198 words in 6 minutes. On his second a

ssignment, he types 248 words in 8 minutes. Did his speed improve? Tell how you know. use unit rate and show work please.
Mathematics
2 answers:
nataly862011 [7]3 years ago
6 0
No he actually got slower. if you divide 198\6 it equals 33 if you divide 248\8 you get 31
RSB [31]3 years ago
3 0
Yes by 2 seconds
198 divided by 6 is 33
258 divided by 8 is 31
And 31 is before 33
So 31 is quicker which means it's
Yes he improved
Btw plz accept my request
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PLEASE HELP!
kirza4 [7]
Equations with absolute value:

|f(x)| = k

Where k is a positive number; if k is a negative number, the equation is impossible (absolute value is always positive).

How to solve:

|f(x) | = k
- f(x) = k \: \: or \: \: f(x) = k

Then:
1. |x+7|=12
x+7=12 V -x-7=12
x=5 V -x=19
x=5 V x=-19
{-19, 5}

2. |2x+4|=8
2x+4=8 V -2x-4=8
2x=4 V -2x=12
x=2 V x=-6

3. 3|3k|=27
3×3k=27 V 3×(-3k)=27
9k=27 V -9k=27
k=3 V k=-3
{-3, 3}

4. 5|b+8|=30
5×(b+8)=30 V 5×(-b-8)=30
5b+40=30 V -5b-40=30
5b=-10 V -5b=70
b=-2 V b=-14
{-14, -2}

5. |m+9|=5
m+9=5 V -m-9=5
m+9=5 V m+9=-5
4 0
3 years ago
Read 2 more answers
#12 please<br> I don't if anyone knows the GUESS method, that would be great
pochemuha
An hour and five minutes
4 0
3 years ago
The length of a rectangle is 2 ft longer than its width.
notsponge [240]
If the length is 2ft longer than it’s width
Let’s assume width as X and so the length will be X+2
Perimeter=40ft
Putting the values into a equation, we get,
X+2+X=40
=2X+2=40
=2X=38
=X=19
Therefore
Length =X+2=19+2
=21ft
Width=X=19ft
Area of rectangle =lxb
= 21x19
3992ft
HOPE IT WAS HELPFUL :)
6 0
3 years ago
A student is trying to solve the system of two equations given below:
erma4kov [3.2K]
Your answer is the third one. WHen it comes to system of equations, this will cancel out 4a, therefore isolating the variable. That turns the equation into -2b=-5.
5 0
3 years ago
Find the absolute maximum and absolute minimum values, if any, of the function. (If an answer does not exist, enter DNE.)f(x) =
lyudmila [28]

Answer:

The absolute minimum value is "-\frac{21}{4}" and the absolute maximum value is "15".

Step-by-step explanation:

Given:

f(x)=x^2-x-5

on,

[0,5]

By differentiating it, we get

⇒ f'(x)=2x-1

Set f'(x)=0

then,

⇒ 2x-1=0

          2x=1

            x=\frac{1}{2} (Critical point)

When x=0,

⇒ f(x)=-5

When x=\frac{1}{2},

⇒ f(x)=-\frac{21}{4} (Absolute minimum)

When x=5

⇒ f(x)=15 (Absolute maximum)

7 0
3 years ago
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