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s344n2d4d5 [400]
3 years ago
8

Which is an equation of the line with x-intercept 2 and y-intercept 12?

Mathematics
1 answer:
Liula [17]3 years ago
5 0
Your answer should be y=2x+12
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Ismail school is at the corner of high Street and second Avenue. The corner forms a right angle. How might he describe the way t
disa [49]

Answer:

Step-by-step explanation:

street never meet

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Find the vertex and length of the latus rectum for the parabola. y=1/6(x-8)^2+6
Ivan

Step-by-step explanation:

If the parabola has the form

y = a(x - h)^2 + k (vertex form)

then its vertex is located at the point (h, k). Therefore, the vertex of the parabola

y = \dfrac{1}{6}(x - 8)^2 + 6

is located at the point (8, 6).

To find the length of the parabola's latus rectum, we need to find its focal length <em>f</em>. Luckily, since our equation is in vertex form, we can easily find from the focus (or focal point) coordinate, which is

\text{focus} = (h, k +\frac{1}{4a})

where \frac{1}{4a} is called the focal length or distance of the focus from the vertex. So from our equation, we can see that the focal length <em>f</em> is

f = \dfrac{1}{4(\frac{1}{6})} = \dfrac{3}{2}

By definition, the length of the latus rectum is four times the focal length so therefore, its value is

\text{latus rectum} = 4\left(\dfrac{3}{2}\right) = 6

5 0
3 years ago
I WILL GIVE Brainiest IF YOUR RIGHT
Brut [27]

Answer:

16.8

Step-by-step explanation:

3 0
2 years ago
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You run 4and3/4 miles through a park each day if you ran that distance five times last week how many miles did you run
zzz [600]
23.75 miles, because you would take 4 and 3/4 (4.75) and multiply by 5 :)
7 0
3 years ago
Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = ln x, [1,7] choose one letter t
inn [45]
ANSWER

B.Yes, f is continuous on [1, 7] and differentiable on (1, 7).

c\approx 3.08

EXPLANATION

The given

f(x) = ln(x)

The hypotheses are

1. The function is continuous on [1, 7].

2. The function is differentiable on (1, 7).

3. There is a c, such that:

f'(c) = \frac{f(7) - f(1)}{7 - 1}

f'(x) = \frac{1}{x}

This implies that;

\frac{1}{c} = \frac{ ln(7) - 0}{6}

\frac{1}{c} = \frac{ ln(7)}{6}

c = \frac{6}{ ln(7) }

c\approx 3.08

Since the function is continuous on [1, 7] and differentiable on (1, 7) it satisfies the mean value theorem.
7 0
3 years ago
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