Eliminate the x and y

Answer:

D or answer choice 4

Step-by-step explanation:

Answer:

y = 8x + 12

Step-by-step explanation:

y = (2x + 3)4

Distribute the 4 to the 2x  and 3

y = 8x + 12

^ Expanded form

Think of it this way, a function must pass the vertical line test.  There may not be 2 outputs for one input for example  (0, 1) (0, 2)  However, you can have something like this (1, 0) (2, 0)

This is indeed a function as it meets those criterion.  There are no two outputs for one input.

Answer:

49

Step-by-step explanation:

If there's 60 beads in total and 11 are round and the rest aren't all you have to do is subtract 11 from 60.

Like This: 60-11=49

To check your answer you can add them to see if you get 60. Like this: 11+49=60

Answer:

3(x+1)

Step-by-step explanation:

Answer:

No

No

Yes

Yes

No

Step-by-step explanation:

In this case we are looking at substitution, so any where you find x put three(3) in replace of it.

And remember the answer should be 36,

I find it difficult to type the answer so I will take a picture instead.

Answer:

10.5

Step-by-step explanation:

Hope I helped please mark brainliest if I did and thanks! <3

Answer:

0.032

Step-by-step explanation:

Answer:  0.03284281007

Step-by-step explanation:

Answer:

  A) 1: exponential; 2: linear

  B) 1: f(x) = 45000·0.9^x; 2: f(x) = -3000x+45000

  C) The difference is significant. 1: $11,348; 2: $6,000

Step-by-step explanation:

The values for Car 1 have a common ratio of 40500/45000 = 0.9. An exponential function is used to describe a sequence where the terms have a common ratio.

The values for Car 2 have a common difference of 42000-45000 = -3000. A linear function is used to describe a sequence where the terms have a common difference.

__

An exponential function can be written as ...

  f(x) = (initial value)·(common ratio)^x

A linear function can be written as ...

  f(x) = (common difference)x + (initial value)

The functions are ...

  Car 1: f(x) = 45000·0.9^x

  Car 2: f(x) = -3000x +45000

___

The formulas for the different car values after 13 years give ...

 Car 1: f(13) = 45000·0.9^13 ≈ 11,348

  Car 2: f(13) = -3000(13) +45000 = 6,000

The value of Car 1 will be almost double the value of Car 2 after 13 years. Yes the difference in value is significant.

Answer:

Step-by-step explanation:

27836?+_68_729+_8978

Answer:

-1

sorry if this is wrong I really am

Answer:

m= -1

Step-by-step explanation:

Have a nice day

Answer:

$6,000

Step-by-step explanation:

$6,000 Because miles younger than 50

Answer A is correct.

Step-by-step explanation:

Negative one is less than one.

Answer:

  $700 for a chair and a table

Step-by-step explanation:

Let t and c represent the price of a table and a chair, respectively. The given relations are ...

  10c = 4t

  15c +2t = $4000

Divide the first equation by 2 and substitute for 2t:

  5c = 2t

  15c +5c = $4000

  20c = $4000

  c = $200

  t = 5c/2 = 5($200)/2 = $500

The total price of a chair and a table is ...

  $200 +500 = $700 . . . . price of a chair and a table

Answer:

A is the cheapest deal.

Step-by-step explanation:

find unit rate; cost / quantity

20.20 / 10 = $2.02 per gallon

26.04 / 12 = $2.17 per gallon

28.24 / 24 = $2.01 per gallon

30.45 / 25 = $2.03 per gallon

9514 1404 393

Answer:

  x = -2

Step-by-step explanation:

Multiply both sides of the equation by -1.

  (-1)(2) = (-1)(-x)

  -2 = x

The value of x that makes the equation true is x = -2.

[tex]W = \displaystyle \int^{5}_{2} F(x) ~ dx\\\\\\= \displaystyle \int^{5}_{2} \left(12x^2 +7\right) ~dx\\\\\\= 12\displaystyle \int^{5}_{2} x^2 ~dx + \displaystyle \int^{5}_{2} 7~ dx\\\\\\=12 \left[\dfrac{x^3}3 \right]^{5}_{2} + 7\left[x\right]^{5}_{2}\\\\\\=4(5^3-2^3) + 7(5-2) = 468 + 21 = 489~ J[/tex]

Answer:

x equals -2 and y equals−4

Answer:

(3*5)+(19*3)

Step-by-step explanation:

both numbers GCF is 3.

(3*5)=15

(19*3)=57

15+57=72

Hope that helps

Answer:

10

Step-by-step explanation:

13 pieses - 3 pieces = 10 pieces that are eaten

Answer:

10

Step-by-step explanation:

Check the picture below.

now, notice the triangles made, 3 right triangles with legs of 5 and 4 each.

looking at quadrilateral (1), from the leg 5 units long, we have a line with a run of -1 and a rise of 3, from the leg 4 units long, we have a line with a run of 4 and a rise of -1.

Now, quadrilaterals (2) and (4) display the same lengths and rise and runs, which means they're all congruent, quadrilateral (3) has no equal.

so 1 ≅ 2 and 1 ≅ 4.

Answer:

319 books

Step-by-step explanation:

345+179=524

524-205 = 319

Answer:

16 I guess.

Step-by-step explanation:

lolololololol?

Answer:

£6.25

Step-by-step explanation:

2.25 x 15 = 33.75

40 - 33.75 = 6.25

Answer:

£6.25

Step-by-step explanation:

Hii,

So if each lesson is 2.25, you times this by 15 to get the total amount of payment. £2.25 x 15 = 33.75

To get change, you need to take away the amount you are giving by how much the bill is.

£40 - £33.75 = £6.25

I hope this helped :P

Answer:

D. is your answer

"Find all real solutions of" basically means solve for x. You're looking for the real numbers x such that the given equation is satisfied. "Zeros" is synonymous with "solutions".

It looks like the first equation is

3x³ + x² - 38x + 24 = 0

It's not immediately obvious whether this can be factorized, so we first carry out the rational zero test to see if there are any rational solutions. *If* this equation has any rational solutions, then they must take the form of (divisor of 24)/(divisor of 3). That is, take any divisor of the constant coefficient and divide it by any divisor of the leading term's coefficient.

We have

• divisors of 24 : ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24

• divisors of 3 : ±1, ±3

Sadly, there are 64 possible roots to test, but we only need one to make the rest of the problem easier.

• Let x = +1 / +1 = 1. Then plugging x = 1 into the equation gives

3•1³ + 1² - 38•1 + 24 = -10 ≠ 0

so x = 1 is *not* a solution

• Let x = +2 / +1 = 2. If x = 2, then

3•2³ + 2² - 38•2 + 24 = -24 ≠ 0

so x = 2 is *not* a solution.

• Let x = +3 / +1 = 3. If x = 3, then

3•3³ + 3² - 38•3 + 24 = 0

so x = 3 is a solution.

By the remainder theorem, this means that x - 3 divides 3x³ + x² - 38x + 24. Using long or synthetic division,

(3x³ + x² - 38x + 24) / (x - 3) = 3x² + 10x - 8

so that

3x³ + x² - 38x + 24 = 0

is the same as

(x - 3) (3x² + 10x - 8) = 0

We factorize the remaining quadratic to get

3x² + 10x - 8 = (3x - 2) (x + 4)

and so the original equation is equivalent to

(x - 3) (3x - 2) (x + 4) = 0

Solve for x :

x - 3 = 0   or   3x - 2 = 0   or   x + 4 = 0

x = 3   or   x = 2/3   or   x = -4

The other equations can be solved similarly, though some of the others you included are much simpler.

For the second problem (note that you left out an exponent), we solve the equation

x³ - 2x² - 23x + 60 = 0

Again, not immediately obvious how to factorize this. But using the rational root test, we have

• divisors of 60 : ±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60

• divisors of 1 : ±1

If you follow the same steps as before, you'll find the first rational solution with x = 3. So x - 3 divides x³ - 2x² - 23x + 60, and long division gives

(x³ - 2x² - 23x + 60) / (x - 3) = x² + x - 20

and the resulting quadratic is easily factored,

x² + x - 20 = (x - 4) (x + 5)

So, we have

x³ - 2x² - 23x + 60 = (x - 3) (x - 4) (x + 5) = 0

which means

x - 3 = 0   or   x - 4 = 0   or   x + 5 = 0

x = 3   or   x = 4   or   x = -5

For the third problem, we can factor by grouping.

x³ - 4x² - x + 4 = 0

x² (x - 4) - (x - 4) = 0

(x² - 1) (x - 4) = 0

(x - 1) (x + 1) (x - 4) = 0

Then

x - 1 = 0   or   x + 1 = 0   or   x - 4 = 0

x = 1   or   x = -1   or   x = 4

For the fourth problem, we use the rational root test again, but twice this time.

x⁴ - 6x³ + 7x² + 6x - 8 = 0

• divisors of -8 : ±1, ±2, ±4, ±8

• divisors of 1 : ±1

We find that x = 1 and x = 2 are both rational roots, so both x - 1 and x - 2 divide x⁴ - 6x³ + 7x² + 6x - 8. By long division,

(x⁴ - 6x³ + 7x² + 6x - 8) / ((x - 1) (x - 2)) = x² - 3x - 4

and factoring this gives

x² - 3x - 4 = (x + 1) (x - 4)

So, we have

x⁴ - 6x³ + 7x² + 6x - 8 = (x - 1) (x - 2) (x + 1) (x - 4) = 0

which means

x - 1 = 0   or   x - 2 = 0   or   x + 1 = 0   or   x - 4 = 0

x = 1   or   x = 2   or   x = -1   or   x = 4

For the last problem,

x⁴ + 4x³ + 7x² + 16x + 12 = 0

we use the rational root test again.

• divisors of 12 : ±1, ±2, ±3, ±4, ±6, ±12

• divisors of 1 : ±1

We find that both x = -1 and x = -3 are rational solutions, and dividing gives

(x⁴ + 4x³ + 7x² + 16x + 12) / ((x + 1) (x + 3)) = x² + 4

which we cannot factorize further over the real numbers. So we end up with

x⁴ + 4x³ + 7x² + 16x + 12 = (x + 1) (x + 3) (x² + 4) = 0

so that

x + 1 = 0   or   x + 3 = 0   or   x² + 4 = 0

x = -1   or   x = -3

(We omit the third equation here because it has no real solution since x² + 4 ≥ 4 for all real x.)