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igor_vitrenko [27]
3 years ago
6

Does anyone know this answer? timed test

Mathematics
2 answers:
omeli [17]3 years ago
3 0
The answer is
(2/1/8)

D
Nadya [2.5K]3 years ago
3 0
The answer is d because when multiplying you car 9 5/24
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Ok so I have 40 and I need to make it into a mixed fraction
slava [35]
Your Answer Is 40 It’s Simple Really If You Kinda Think About It But Has A Fraction It Would Be 40/1
8 0
3 years ago
Mathia question order or operations
eduard

Answer:

53

Step-by-step explanation:

Use pemdas

Use P and do (2+2)

then E 2^3 to get 8

Now you have 9-4/4+8*5

Do M and D To get 9-1+45

Now add and subtract to get your answer

7 0
3 years ago
HELP ME QUICK PLEASEE!!
Lena [83]

Answer:

  A) is a function

  B) f(x) is greater

  C) x = 12

Step-by-step explanation:

A) Yes, the relation is a function.

A function has a unique input value for each output value. There are no repeated x-values, so this relation is a function.

__

B) The table says y=6 for x=6.

  f(6) = 2(6)+16 = 28

f(x) has a greater value for x=6.

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C) Substituting the given information, we have ...

  40 = 2x +16

  24 = 2x

  12 = x

The value of x is 12 when f(x) is 40.

3 0
3 years ago
The town librarian bought a combination of new-release movies on DVD for $20 and classic movies on DVD for $8.
blagie [28]

Answer:

16 new releases and 22 classic movies.

8 0
1 year ago
Hi, the answer is K but can anyone show why
dlinn [17]

Let's do 51 and 52.

51. The contrapositive has the same truth value as the original statement. That's opposed to the converse, which may or may not be true independent of the original statement.

The contrapositive of IF P THEN Q is IF not Q THEN not P.  They're equivalent.  Here that's If the cat is not female then it is not tricolor.

Answer: C

52.  

(x^3)^{(4-b^2)}=1

x^{3(4-b^2)} = 1

For the statement to be true, the exponent must be zero:

3(4-b^2) = 0

b^2 = 4

b = \pm 2

Both positive 2 and negative 2 have a square of 4.

Answer: K

By the way, usually we assume 0^0=1 so the restriction that x \ne 0 isn't really necessary.  Think of the definition of a polynomial or the binomial expansion:

\displaystyle f(x)=\sum_{k=0}^n a_k x^k

\displaystyle(x+y)^n=\sum_{k=0}^n {n \choose k} x^{k}y^{n-k}

For these common equalities to work when x=0 we need to define 0^0=1


3 0
3 years ago
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