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sweet-ann [11.9K]
3 years ago
13

Consider the quadratic function:

Mathematics
1 answer:
sdas [7]3 years ago
8 0

Answer:

Step-by-step explanation:

I'm not sure how you are finding this in class, but the easiest way by far is to use the expressions for h and k:

h=\frac{-b}{2a}  and

k=c-\frac{b^2}{4a}

(Just a fun fact:  Those expressions come from the quadratic formula that help us to factor a quadratic equation.)

Filling in for h:

h=\frac{-(-8)}{2(1)}  simplifies to

h = \frac{8}{2}=4

Filling in for k:

k=-9-\frac{(-8)^2}{4(1)}  which simplifies a bit to

k=-9-\frac{64}{4}  which simplifies a bit more to

k=-9-16=-25

The vertex, then, is (4, -25).

You would also find this if you completed the square.  But again, I'm not sure how you're solving for the vertex in class.

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Given: ABCD trapezoid<br> BC=1.2m, AD=1.8m<br> AB=1.5m, CD=1.2m<br> AB∩CD=K<br> Find: BK and CK
stepladder [879]

Answer:

The value of BK is 3m and the value of CK is 2.4m.

Step-by-step explanation:

Given information: ABCD trapezoid

, BC=1.2m, AD=1.8m

, AB=1.5m, CD=1.2m

, AB∩CD=K.

Using the given information draw a figure.

Two sides of a trapezoid are parallel.

Since AB∩CD=K, therefore AB and CD are not parallel, because parallel line never intersect.

AD\parallelBC

\angle KBC=\angle KAD              (Corresponding angles)

\angle KCB=\angle KDA             (Corresponding angles)

By AA rule of similarity

\triangle KBC\sim \triangle KAD

Corresponding sides of similar triangles are proportional.

\frac{KB}{KA}=\frac{KC}{KD}=\frac{BC}{AD}

\frac{x}{x+1.5}=\frac{y}{y+1.2}=\frac{1.2}{1.8}

\frac{x}{x+1.5}=\frac{1.2}{1.8}

\frac{x}{x+1.5}=\frac{2}{3}

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x=3

The length of BK is 3 m.

\frac{y}{y+1.2}=\frac{1.2}{1.8}

\frac{y}{y+1.2}=\frac{2}{3}

3y=2y+2.4

y=2.4

The length of CK is 2.4 m.

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3 years ago
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