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Oksanka [162]
2 years ago
12

What is the difference? Complete the equation. -1 2/5 - (-4/5) = ?

Mathematics
1 answer:
AfilCa [17]2 years ago
7 0

Answer:

First convert them which will be

-7/5 - (-4/5)

so when you subtract a negative number from negative number they actually subtract ex = -4-(-2) = -2

so its simply 7/5-4/5 then add a negative sign

so

3/5

now add negative sign so

-3/5

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Effectus [21]

Answer:

Yes, this function can be defined by linear model.

Step-by-step explanation:

\mathrm{Total\:=\:Fixed\:Fee\:+\:Hourly\:Rate\:\times Number\:of\:hours}

\mathrm{Linear:}\:\:T=\:$25+30h

Best Regards!

3 0
2 years ago
Need help ASAP! Show work if you can, it’s fine if not tho!
Elza [17]

Answer:

Step-by-step explanation:

(2(6) - 3(10)^2)/(|6 - 10)|

(12 - 300)/|-4|

-288/4 = -72

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3 0
2 years ago
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Which property is demonstrated by following statement 16+(22+a)=16+(a+22)
beks73 [17]

commutative property

3 0
3 years ago
(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

7 0
3 years ago
Solve for x: negative 7 over 6, multiplied by x minus 6 equals negative 48
Usimov [2.4K]

-7/6*x-6=-48

X will equal -36.

7 0
3 years ago
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