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ss7ja [257]
3 years ago
15

David had $30 to spend on three gifts.He spent 10 ¼ dollars on gifts A and 5 ⅘ dollars on gifts B.How much money did he have lef

t for gifts C.
Mathematics
1 answer:
Katen [24]3 years ago
6 0

Answer: $13.95

Step-by-step explanation:

1/4 of a Dollar is 25 cents

4/5 of a Dollar is 80 cents

30 - 10.25 = 19.75

19.75 - 5.80 = 13.95

David has $13.95 to spend of Gift C

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Week 1-2 so it would be week one
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3 years ago
What are the independent and dependent variables in the relationship?
Sindrei [870]

w is the width and it is the independent variable. You can pick any positive whole number you want (within reason of course; you can't go to infinity or go beyond some set boundary). Whatever you picked for w, the expression 2w-5 will be dependent on it. So the length is dependent on the width.

For instance, if the width is w = 10 feet, then 2w-5 = 2*10-5 = 20-5 = 15 feet is the length. The choice of 10 feet for the width directly affects the length being 15 feet.

5 0
3 years 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
4 years ago
7 cards are drawn from a standard deck of 52 playing cards. How many different 7-card hands are possible if the drawing is done
Vladimir [108]

Answer:

The number of different 7 card hands possibility is  133784560

Step-by-step explanation:

The computation of the number of different 7 card hands possibility is shown below:

Here we use the combination as the orders of choosing it is not significant

So,

The number of the different hands possible would be

= ^{52}C_7

= 52! ÷ (7! × (52 - 7)!)

= 133784560

Hence, the number of different 7 card hands possibility is  133784560

4 0
3 years ago
3
Juli2301 [7.4K]
A 4.2%
Explanation: that might be the answer ✌︎☆♪ ٩( ᐛ )و
8 0
2 years ago
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