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Fantom [35]
4 years ago
5

Write an algebraic expression for the word phrase. five fewer than d days​

Mathematics
1 answer:
kogti [31]4 years ago
3 0

d-5

because it is 5 fewer we are subtracting the number five from the variable representing days.

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How do I solve 7x+3=59​
lesya692 [45]

Answer:7x+3=59

59-3

7x=56

Then you divide them both by 7

7x/7=56/7

Then you get your answer

X=?

I'll let you do the last part tho

Step-by-step explanation:

7 0
3 years ago
If I paid $9.00 for 4 lb of candies, how much candies can I buy for $12.00?
Lina20 [59]

Answer:

We are given that 4 lb of candies are bought for $9.00

Therefore, the price of 1 lb of candies is:

\frac{9}{4}=2.25

Now the amount of candies that can bought for $12.00 is:

\frac{12}{2.25} =5.3333 lb

Therefore, 5.3333 lb of candies can be bought for $12.00


3 0
3 years ago
Rewrite the equation of the parabola in vertex form.
aleksandrvk [35]
The answer would be y=(x+4)^2-37
5 0
3 years ago
Find the greatest common factor of 15y^3 and 9n^4​
STALIN [3.7K]

The answer is 3. The only common factor between the numbers is 3, because 15 is 3 times 5, while 9 is 3 times 3.

8 0
3 years ago
Read 2 more answers
15 players for a softball team show up for a game: (a) How many ways are there to choose 10 players to take the field? 3003 (b)
prohojiy [21]

Answer:

(a) 3003 ways

(b) 10897286400 ways

(c) 3002 ways

Step-by-step explanation:

Given

n = 15 --- 15 players

Solving (a) Ways of selecting 10 players.

This implies combination.

So, we have:

r=10

Using:

^nC_r = \frac{n!}{(n-r)!r!}

We have:

^{15}C_{10} = \frac{15!}{(15-10)!10!}

^{15}C_{10} = \frac{15!}{5!10!}

Simplify

^{15}C_{10} = \frac{15*14*13*12*11*10!}{5!10!}

^{15}C_{10} = \frac{15*14*13*12*11}{5!}

^{15}C_{10} = \frac{15*14*13*12*11}{5*4*3*2*1}

^{15}C_{10} = \frac{360360}{120}

^{15}C_{10} = 3003

Solving (b) Ways of assigning positions to 10 players.

This implies permutation.

So, we have:

r=10

Using:

^nP_r = \frac{n!}{(n-r)!}

We have:

^{15}P_{10} = \frac{15!}{(15-10)!}

^{15}P_{10} = \frac{15!}{5!}

Solve each factorial

^{15}P_{10} = \frac{1307674368000}{120}

^{15}P_{10} = 10897286400

Solving (c) Ways of choosing at least 1 woman

We have:

Men = 10

Women = 5

Ways of selecting 10 players is: (a) 3003 ways

Since the number of men are 10, there is 1 way of selecting 10 men (i.e. selection without women)

Using complement rule:

At least 1 woman = Total - No woman

At\ least\ 1\ woman = 3003 - 1

At\ least\ 1\ woman = 3002

5 0
3 years ago
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