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2 years ago

Use the cubic model y = 5a3 - 2a2 + a - 45 to find the value of y when x = 4.

1 answer:

3 0

Assuming 5a3 equals 5a^3, and the same for everything else, you would plug in 4 for a, so 5(4)^3 - 2(4)^2 + 4 - 45, which equals 247.

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Write 0.9 as a fraction with a denominator of 10

adelina 88 [10]

The answer is 9/10
the denominator is on the bottom

5 0

2 years ago

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How many arrangements can be made using 2 letters of the word HYPERBOLAS if no letter is to be used more than once?

alisha [4.7K]

Answer:

90 arrangements

Step-by-step explanation:

Since there are no repititions of letters, there are unique 10 letters in total.

THe number of arrangements would be 2 permutation 10. We need the formula for permutation. That is:

$nPr=\frac{n!}{(n-r)!}\\$

Now, n = 10 [total] and r is 2, so we have:

$nPr=\frac{n!}{(n-r)!}\\\\_{10}P_{2}=\frac{10!}{(10-2)!}\\=\frac{10!}{8!}\\=\frac{10*9*8!}{8!}\\=10*9\\=90$

So, there can be 90 arrangements

3 0

3 years ago

AB = 3 + x

strojnjashka [21]

Answer:

Option A) 4, 2

Step-by-step explanation:

We are given that quadrilateral ABCD is a parallelogram with sides:

$AB = 3 + x\\DC = 4x\\AD = y + 1\\BC = 2y$

Since the opposite sides of a parallelogram are equal, we can write:

$AB = DC\\AD=BC$

Equating the sides we get,

$\Rightarrow 3+x = 4x\\4x-x = 3\\3x = 3\\x = 1\\\Rightarrow y + 1 = 2y\\y = 1$

Putting the values of x and y, we get:

$AB = 3 + x =4\\DC = 4x =4\\AD = y + 1=2\\BC = 2y=2$

Thus, the lengths of the opposite side pairs is 4,2.

4 0

2 years ago

Carson's wheelbarrow can hold 345 pounds. If he has 121 pound of rock in the wheelbarrow what number of pounds p can he pull wit

dangina [55]

Answer:

$p\leq 224\ lb$

The maximum number of pounds p is 224

Step-by-step explanation:

Let

p ----> the maximum number of pounds he can pull without going over the weight limit

we know that

121 pound of rock in the wheelbarrow plus p pounds must be at most 345 pounds

In this problem the word "at most" means "less than or equal to"

The inequality that represent this situation is

$p+121\leq 345$

solve for p

subtract 121 both sides

$p\leq 345-121$

$p\leq 224\ lb$

therefore

The maximum number of pounds is 224

7 0

3 years ago

Is this right??? Please answer

lesya692 [45]

Answer:Yes its correct!(good job)

Step-by-step explanation:If you want to doubule check just count the tens carefully and if your still sketchy about it just cir cle the ones you counted hope this helped!

7 0

3 years ago

Read 2 more answers