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3 years ago

Determite weather the ordered pair is a solution of the linear equation x=y+1/2;(-5,-2)

Mathematics

1 answer:

Minchanka [31]3 years ago

7 0

Answer:

(y+1/2 ; y)

Step-by-step explanation:

hello : all ordered pair is a solution of the linear equation x=y+1/2;(-5,-2) is :

(y+1/2 ; y)

(-5 ; -2) is not solution because y = -2 but -2+1/2 = -3/2 no -5

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Can someone please help with these 5 questions?<br><br>10 points!

babymother [125]

Answer:

Step-by-step explanation:

1) true

2)3/4x-6= 12 = true

3) 8= -2x + 4 = false

4) x+5 = -3 = false

5) -3= 5-2x = true

7 0

3 years ago

Help meh?

zloy xaker [14]

X= # of miles
.55x+1.75 is greater than or equal to 10

So if she has to go more than 2 miles we will put 2 in for X

.55(2)+1.75 is greater than or equal to 10
.55×2= 1.10+1.75= 2.85
2.85 is greater than 10 is false so now we know she has to go more than 2 miles.

Let's guess and check. Now let's say 15 miles
.55×15= 8.25+1.75= 10
10=10 So she needs to drive at least 15 miles

x is greater than or equal to 15

Please don't solely rely on my answer I don't know if it's correct

3 0

3 years ago

Express this to single logarithm

zzz [600]

Answer: $\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right)$

We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.

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Explanation:

It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.

I'm going to use these three log rules, which apply to any base.

log(A) + log(B) = log(A*B)
log(A) - log(B) = log(A/B)
B*log(A) = log(A^B)

From there, we can then say the following:

$\frac{1}{2}\log_{2}\left(m\right)-3\log_{2}\left(n\right)+2\log_{2}\left(q\right)\\\\\log_{2}\left(m^{1/2}\right)-\log_{2}\left(n^3\right)+\log_{2}\left(q^2\right) \ \text{ .... use log rule 3}\\\\\log_{2}\left(\sqrt{m}\right)+\log_{2}\left(q^2\right)-\log_{2}\left(n^3\right)\\\\\log_{2}\left(\sqrt{m}*q^2\right)-\log_{2}\left(n^3\right) \ \text{ .... use log rule 1}\\\\\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right) \ \text{ .... use log rule 2}$

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2 years ago

0.07 is one tenth of

mr Goodwill [35]

0.07 is one tenth of 0.7

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3 years ago

2/3 of the cards are pitchers and 3/4 of the remainder are outfielders. If the 3 cards left over are infielders, how many cards

snow_tiger [21]

Answer:

2/3 + 3/4 = 1 52/125. That plus 3 is 4 / 52/125. Sorry if this isn't the right answer.

6 0

3 years ago