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3 years ago

Solve (x-2)^2-3=1 graphically.

Mathematics

1 answer:

zimovet [89]3 years ago

8 0

we are given

$(x-2)^2-3=1$

we can seperate them

First function is

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

second function is

$y=1$

now, we can draw graph

and find intersection points

so, we can see that

intersection points are

(0,1) and (4,1)

so, solutions are

$x=0,x=4$ ...........Answer

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Artemon [7]

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2 years ago

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3 years ago

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a utility worker is 5.5 feet tall and is casting a shadow 4 feet long at the same time a nearby utility pole casts a shadow 20 f

Assoli18 [71]

So this is essentially a proportion problem. They want you to be able to recognize a relationship between the two. The easiest way to go about this is to draw a picture (I have included one). What you want to do is notice that you are given both shadow lengths. This is essentially going to be your first step to creating the proportion.

Understanding that we know the shadow lengths, we can turn these into a fraction such as so:

4ft/20ft.

Now that we know that, we need to look at are unknown. Our unknown is the height of the utility pole. We can solve this by substituting for X since we know the utility worker is 5.5 ft tall.

Your fraction for that would look like:

5.5ft/X

Make sure to always make your top to top match, such as since I put the utility worker's shadow on top, I need to put his height on top.

Now we can solve.

4ft/20ft =5.5ft/X

We need to cross multiply to get an equation we can work with. If we cross multiply, Your equation will look like:

4x = 110ft

This is a simple one step equation. Divide both sides by 4 to get your answer.

x= 27.5ft.

This means your Utility pole's height is 27.5ft.

6 0

3 years ago

The equation 12x² + 4kx + 3 = 0 has real and equal roots, if

sleet_krkn [62]

Answer:

$k=\pm3$

Step-by-step explanation:

[...] if you can write the LHS as a perfect square, or if you can't spot a factorization of it right away, if and only if the discriminant $\Delta = b^2-4ac$ (or, if b is an even number, 1/4 of it) is zero.

Stare at it for a while. First term is $3(2x)^2$ , third term is $3(1)^2$ , we are missing a double product, but we can play with k. For the LHS to be $3(2x\pm1)^2 = 3(4x^2\pm4x+1) = 12x^2\pm12x+3$ you just need $4k= \pm12 \rightarrow k=\pm3$ .