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3 years ago

8

Three of the four vertices (corners) of a parallelogram are (-2,7), (3,7) and (2,5). What is the ordered pair of the missing ver

tex (corner)?

2 answers:

Cerrena [4.2K]3 years ago

4 0

(-3, 5) i think, this is just a guess and I haven’t done this stuff since 8th grade mb

Leno4ka [110]3 years ago

3 0

Answer:

It's (-3,5) I got it right .

Step-by-step explanation:

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Find the value of 8!<br> O A. 56<br> O B. 16,777,216<br> O c. 40,320<br> O D. 64

Ksivusya [100]

8! is 40,320

Step-by-step explanation:

Step 1: Find 8! (8 factorial).

Factorial of 8 is the product of 8 and all the integers below 8.

⇒ 8! = 8×7×6×5×4×3×2×1

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3 years ago

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cupoosta [38]

Answer:

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Step-by-step explanation:

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The vertex of the parabola below is at the point (3,2) and the point (4,6) is on the parabola. What is the equation of the parab

nata0808 [166]

Answer:

$\large\boxed{y=4(x-3)^2+2\ \bold{vertex\ form}}\\\boxed{y=4x^2-24x+38\ \bold{standard\ form}}$

Step-by-step explanation:

The vertex form of a parabola:

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

(h, k) - vertex

We have the vertex at (3, 2) → h = 3 and k = 2.

Substitute:

$y=a(x-3)^2+2$

The point (4, 6) is on athe parabola. Put the coordinates of this point to the equation:

$6=a(4-3)^2+2$ <em>subtract 2 from both sides</em>

$6-2=a(1)^2+2-2$

$4=a\to a=4$

Finally:

$y=4(x-3)^2+2$ <em>vertex form</em>

<em>use (a - b)² = a² - 2ab + b²</em>

$y=4(x^2-6x+9)+2$ <em>use the distributive property</em>

$y=4x^2-24x+36+2$

$y=4x^2-24x+38$ <em>standard form</em>

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