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Troyanec [42]
3 years ago
8

I know...another one

Mathematics
1 answer:
asambeis [7]3 years ago
4 0

I'm so confused what do you need help with?

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Write an expression for the verbal phrase: The difference between x and half of y.
zmey [24]

Answer:

C.

Step-by-step explanation:

half of y = 1/2 *y

Difference between x and half of y

= x - 1/2 y

4 0
3 years ago
Graph a line parallel to y-2x+1 and has a y-intercept of (0,5))
Debora [2.8K]

your answer is x = 6

desmos

7 0
3 years ago
Explain your answer please.
andriy [413]
Just do length x width x height
7 0
3 years ago
Identify the vertex, focus, and directrix. y=1/8y^2
Brut [27]
If all the equations for the directrix are "x = " lines then this is a y^2 parabola.  The actual equation is x= \frac{1}{8}y^2.  The standard form for a positive sideways-opening parabola is  (y-k)^2=4p(x-h).  We know from the equation that the vertex of the parabola is at the origin, or else the translation would be reflected within the parenthesis in the equation.  Our equation has no parenthesis to indicate movement from the origin.  The vertex is (0, 0).  Got that out of the way.  That simplifies our standard form down to  y^2=4p(x).  Let's take a look at our equation now.  It is  x= \frac{1}{8}y^2.  We could rewrite it and make it a closer match to the standard form if we multiply both sides by 8 to get rid of the fraction.  That gives us an equation that looks like this:  y^2=8x.  That means that 4p = 8, and p = 2.  p is the distance that the focus and the directrix are from the vertex.  Since this is a positive parabola, it opens up to the right.  Which means, then, that the focus is to the right of the vertex, 2 units to be exact, and the directrix is 2 units to the left of the vertex.  The formula for the focus is (h + p, k).  Our h is 0, our k is 0 and our p is 2, so the coordinates of the focus are (2, 0).  Going 2 units to the left of the origin then puts our directrix at the line x = -2.  Your choice then as your answer is b. 
6 0
3 years ago
In triangle $ABC$, let angle bisectors $BD$ and $CE$ intersect at $I$. The line through $I$ parallel to $BC$ intersects $AB$ and
Umnica [9.8K]

Answer:

41

Step-by-step explanation:

If you work through a series of obscure calculations involving area and the radius of the incircle, they boil down to a simple fact:

... For MN║BC, perimeter ΔAMN = perimeter ΔABC - BC = AB+AC

.. = 17+24 = 41

_____

Wow! Thank you for an interesting question with a not-so-obvious answer.

_____

<em>A little more detail</em>

The point I that you have defined is the incenter—the center of an inscribed circle in the triangle. Its radius is the distance from I to any side, such as BC, for example.

If we use "Δ" to represent the area of the triangle and "s" to represent the semi-perimeter, (AB+BC+AC)/2, then the incircle has radius Δ/s. The area Δ can be computed from Heron's formula by ...

... Δ = √(s(s-a)(s-b)(s-c)) . . . . where a, b, c are the side lengths

For this triangle, the area is Δ = √38480 ≈ 196.1632 units². That turns out to be irrelevant.

The altitude to BC will be 2Δ/(BC), so the altitude of ΔAMN = (2Δ/(BC) -Δ/s). Dividing this by the altitude to BC gives the ratio of the perimeter of ΔAMN to the perimeter of ΔABC, which is 2s.

Putting these ratios and perimeters together, we get ...

... perimeter ΔAMN = (2Δ/(BC) -Δ/s)/(2Δ/(BC)) × 2s

... = (2/(BC) -1/s) × BC × s = 2s -BC

... perimeter ΔAMN = AB +AC

8 0
3 years ago
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