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Blizzard [7]
3 years ago
12

Write an equation for the line parallel to y = –2x – 5 that contains p(–8, 4).

Mathematics
1 answer:
klemol [59]3 years ago
8 0
The answer is
y=-2x+20
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Given a quadratic with one zero at 3 and a vertex at (-1,4), what is the other zero?
s2008m [1.1K]

Answer:

f (x) = a(x - h)2 + k, where (h, k) is the vertex of the parabola. FYI: Different textbooks have different interpretations of the reference "standard form" of a quadratic function. Some say f (x) = ax2 + bx + c is "standard form", while others say that f (x) = a(x - h)2 + k is "standard form".

Step-by-step explanation:

6 0
3 years ago
Please help asap<br><br> y= ?
Tanzania [10]

Answer:

y = 2|x| - 3

Step-by-step explanation:

We'll start with our parent function which is y = |x|. Notice that graph is translated down by 3 units. So, we'll add that:

y = |x| - 3

Also notice that the ordered pair (1, -1) is on the graph. |1| = 1, so a translation of -3 should take it to -2 and it should have been (1, -2). But, it's not so it must be that the graph is stretched/compressed. If you multiply every point by 2, it gives us the result we want. Say for example, f(1) = 2|1| - 3 = 2 - 3  = -1. We get the point (1, -1) which is exactly what we want. We write compressions as (where a is the compression/stretching constant):

y = a|x| + d

Our compression constant is 2, so our final answer is:

<em>y = 2|x| - 3</em>.

5 0
3 years ago
hi! just remember that you are amazing and everything will be okay! never give up! you got this just take a deep breath and reme
ad-work [718]

Answer:

thank you for this inspiration

Step-by-step explanation:

4 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Clarge%7B%5Cbold%20%5Cred%7B%20%5Csum%20%5Climits_%7B8%7D%5E%7B4%7D%20%7Bx%7D%5E%7B2%7D%2
kiruha [24]

Answer:

No solution

Step-by-step explanation:

We have

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

For the sum it is not correct to assume

$\sum_{x=8}^{4}x^2= 8^2 + 7^2+6^2+5^2+4^2 = 64+49+36+25+16 = 190$

Note that for

$\sum_{x=a}^b f(x)$

it is assumed a\leq x \leq b and in your case \nexists x\in\mathbb{Z}: a\leq x\leq b for a>b

In fact, considering a set S we have

$\sum_{x=a}^b (S \cup \varnothing) = \sum_{x=a}^b S + \sum_{x=a}^b \varnothing $ that satisfy S = S \cup \varnothing

This means that, by definition \sum_{x=a}^b \varnothing = 0

Therefore,

$\sum_{x=8}^{4}x^2 = 0$

because the sum is empty.

For

9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)

we have other problems. Actually, this case is really bad.

Note that \cos^2(\infty) has no value. In fact, if we consider for the case

$\lim_{x \to \infty} \cos^2(x)$, the cosine function oscillates between [-1, 1] , and therefore it is undefined. Thus, we cannot evaluate

9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)

and then

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

has no solution

7 0
3 years ago
1. Graph the function g(x) = [X1 + 4 as a translation of the
Elenna [48]

Answer:

The translation did not effect the domain

The translation effected the range

Step-by-step explanation:

The given graph represents f(x) = IxI

The domain of f(x) is {x : x ∈ R}, where R is the set of real numbers

The range of f(x) is {y : y ≥ 0}

The graph of g(x) is attached down

g(x) = IxI + 4

The domain of g(x) is {x : x ∈ R}, where R is the set of real numbers

The range of g(x) is {y : y ≥ 4}

The graph of g(x) is the image of the graph of f(x) after translate it 4 units up

∵ The translation is vertically

∴ The domain of f(x) = the domain of g(x)

∵ The domain of f(x) is {x : x ∈ R}

∴ The domain of g(x) is {x : x ∈ R}

∴ The translation did not effect the domain

∵ g(x) = f(x) + 4

∵ The range of f(x) is {y : y ≥ 0}

- Add 4 to the value 0

∴ The range of g(x) is {y : y ≥ 0 + 4}

∴ The range of g(x) is {y : y ≥ 4}

∴ The range of g(x) not equal the range of f(x)

∴ The translation effected the range

5 0
3 years ago
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