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2 years ago

-7x+4y=24, 4x-4y=0 substitution

1 answer:

8 0

The first one: -7(x) + 4(y) = 24
The second one: 4(x) - 4(y) = 0

4y - 7x = 24 which is reduced to a negative
(-4y + 4x = 0 )

4(x) = 4(y)

-7(y ) + 4(y) = 24
- 3(y) = 24

3(y) = - 24

y-intercept: - 8
x-intercept: -8

The X & Y intercept is: -8, -8

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How to do this question plz answer

harina [27]

9514 1404 393

Answer:

300

Step-by-step explanation:

There are 25 ways to select the first student. After that student is removed from the selection pool for the second student, there are 24 ways to select the second student. This gives 25·24 = 600 ways to select 2 students in a particular order.

Since we don't care about the order, we can divide this number by the number of ways two students can be ordered: AB or BA, 2 ways.

600/2 = 300

There are 300 ways to pick a combination of two students from 25.

Additional comments

This sort of selection (2 out of 25) has a formula for it, and an abbreviation for the formula.

"n choose k" can be written nCk or C(n, k)

The function is a ratio of factorials:

nCk = n!/(k!(n-k)!)

If you can typeset this, it is written ...

$\displaystyle\binom{n}{k}=\frac{n!}{k!\cdot(n-k)!}$

This is different from the formula for the number of permutations of n things taken k at a time. That would be written nPk or P(n, k) = n!/(n-k)!.

5 0

2 years ago

Read 2 more answers

Which of the following are solutions to the quadratic equation? Check all that

Firlakuza [10]

Answer:

$\boxed{\sf \ \ \ x=-9 \ \ or \ \ x=2 \ \ \ }$

Step-by-step explanation:

hello,

$2x^2+7x-14=x^2+4\\ 2x^2+7x-14-x^2-4=0\\ x^2+7x-18=0\\x^2-2x+9x-18=0\\ x(x-2)+9(x-2)=0\\ (x+9)(x-2)=0\\ x+9 = 0 \ \text{or} \ x-2=0\\ x = -9 \ \text{or} \ x=2$

hope this helps

6 0

2 years ago

16% of 130 is what number

earnstyle [38]

130 • .16 = 20.8 or 20 (4/5) or 104/5
Put the answer as 20.8 unless specified otherwise.
.16 is 16%

5 0

3 years ago

Read 2 more answers

Could you please help with this question for domain, range, and end behavior? I started it, but I think I'm doing it wrong.

Tanzania [10]

It looks like you have the domain confused for the range! You can think of the domain as the set of all "inputs" for a function (all of the x values which are allowed). In the given function, we have no explicit restrictions on the domain, and no situations like division by 0 or taking the square root of a negative number that would otherwise put limits on it, so our domain would simply be the set of all real numbers, R. Inequality notation doesn't really use ∞, so you could just put an R to represent the set. In set notation, we'd write

$\{x\in\re\mathbb{R}\}$

and in interval notation,

$(-\infty,\infty)$

The range, on the other hand, is the set of all possible outputs of a function - here, it's the set of all values f(x) can be. In the case of quadratic equations (equations with an x² term), there will always be some minimum or maximum value limiting the range. Here, we see on the graph that the maximum value for f(x) is 3. The range of the function then includes all values less than or equal to 3. As in inequality, we can say that

$f(x)\leq3$ ,