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ANTONII [103]
2 years ago
9

In the diagram EF is the bisector of <DEF = m < D' EF.

Mathematics
1 answer:
finlep [7]2 years ago
4 0

hii! i just wanted to say hope you're having a fantastic day and hope that things are going well (: stay strong and stay safe!

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-8=-2(z+7) [solving linear equations ]
Sveta_85 [38]
Z=-3 is the correct answer. Use math papa if you need help with the steps on how we got z=-3
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Someone please give me the answers for this lol​
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A is the correct answer.

Step-by-step explanation:

y and x meet on those numbers. And i did the test.

hope this helps

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Help I will mark brainly
Schach [20]

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it is -25/18

Step-by-step explanation:

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Compare the triangles and determine whether they can be proven congruent by SSS, SAS, ASA, AAS, or HL. If not, type "NONE".
EleoNora [17]

Answer: HL

Step-by-step explanation:

The hypotenuse is actually a shared side of the two triangles. (The line segment that crosses the middle of the rectangle) The hypotenuse is the side of a triangle opposite the 90 degree angle. We can use this for hypotenuse congruency. The diagram also shows the shorter legs of the two triangles are congruent. (The legs refer to the sides that make the 90 degree angle)

So we have hypotenuse congruency and leg congruency.

=HL

6 0
2 years ago
Derive the equation of the parabola with the focus <br> is (-7,5) and the directrix of y=-11
Volgvan
So, notice, the focus point is at -7, 5, and the directrix is at y = -11.

keep in mind that the vertex is half-way between those two fellows, and the distance from the vertex to either one of them is "p" units, check the picture below.

with that focus point and that directrix, the half-way over the axis of symmetry will be -7, -3, that's where the vertex is at, and notice the distance "p", is 8 units.

since the parabola is opening upwards, "p" is positive 8.

\bf \textit{parabola vertex form with focus point distance}\\\\&#10;\begin{array}{llll}&#10;4p(x- h)=(y- k)^2&#10;\\\\&#10;\boxed{4p(y- k)=(x- h)^2}&#10;\end{array}&#10;\qquad &#10;\begin{array}{llll}&#10;vertex\ ( h, k)\\\\&#10; p=\textit{distance from vertex to }\\&#10;\qquad \textit{ focus or directrix}&#10;\end{array}\\\\&#10;-------------------------------\\\\&#10;\begin{cases}&#10;h=-7\\&#10;k=-3\\&#10;p=8&#10;\end{cases}\implies 4(8)[y-(-3)]=[x-(-7)]^2&#10;\\\\\\&#10;32(y+3)=(x+7)^2\implies y+3=\cfrac{1}{32}(x+7)^2&#10;\\\\\\&#10;y=\cfrac{1}{32}(x+7)^2-3

8 0
3 years ago
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