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mariarad [96]
3 years ago
13

If f(2) =20 and f(3) =10F(4) = ,f(5)=

Mathematics
1 answer:
Arada [10]3 years ago
5 0
F(4)=0 and F(5)= -10
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The graph below shows the number of shirts of each color in a store. How many more red shirts than the total number of blue shir
MArishka [77]
The answer is A) 15, Total blue and yellow shirts is 30 were red is 45

45-30=15
4 0
3 years ago
24y - 22 = 4 ( 6y - 6 )
Pie
I don’t think their is a solution to this equation
because if you expand the second half it is= 24y-24 which would make the equation
- 24y-22=24y-24
and because the number next to the y is the same on both sides, no matter what y is if we subtract different numbers from each side we will never get the same value for each side of the =
5 0
3 years ago
This is a proportion. What is the value of X?
Verizon [17]

Answer:

x = 15 or x = - \frac{2}{5}

Step-by-step explanation:

Cross- multiplying gives

(14x + 6)(17x + 5) = 9x(27x + 11) ( expanding factors )

238x² + 172x + 30 = 243x² + 99x

rearrange into standard form : ax² + bx + c = 0

5x² - 73x - 30 = 0 ← in standard form

consider the factors of the product 5 × - 30 = - 150 which sum to the coefficient of the x-term (- 73 )

the factors are - 75 and + 2

Use these factors to split the middle term

5x² - 75x + 2x - 30 = 0 ( factor by grouping )

5x(x - 15) + 2(x - 15) = 0 ← take out the factor (x - 15)

(x - 15)(5x + 2) = 0

equate each factor to zero and solve for x

x - 15 = 0 ⇒ x = 15

5x + 2 = 0 ⇒ x = - \frac{2}{5}


7 0
3 years ago
Can someone please solve this
mote1985 [20]
That’s pretty easy you just have to basically y=mx+b
5 0
2 years ago
for the given license plate configuration, determine how many different plates are possinle if letters and digits (a) can ba rep
Inessa05 [86]

Answer:

a) 1,757,600 different plates possible.

b) 1,404,000 different plates possible.

Step-by-step explanation:

The permutations formula is important to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

P_{(n,x)} = \frac{n!}{(n-x)!}

(a) can be repeated:

There are 3 letters, each one with 26 possible outcomes.

2 digits, each with 10 possible outcomes

So

T = 26*26*26*10*10 = 1757600

1,757,600 different plates possible.

(b) cannot be repeated:

Here the permutations formula is used.

Three letters, from a set of 26.

Two digits, from a set of 10. So

T = P_{(26,3)}*P_{(10,2)} = \frac{26!}{(26-3)!}*\frac{10!}{(10-2)!} = 1404000

1,404,000 different plates possible.

4 0
3 years ago
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