Help pls!! I don’t understand

How do you this question? Please explain.

Nuetrik [128]

Answer:

(C) 2√15

Step-by-step explanation:

Recognize that all the triangles are right triangles, so are similar to each other. In these similar triangles, the ratio of the short side to the long side is the same for all.

... CB/CA = CT/CB

... CB² = CA·CT = 10·6 = 60 . . . . . . . . . . multiply by CA·CB; substitute values

... CB = √60 = 2√15 . . . . . . . take the square root; simplify

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<em>Comment on this solution</em>

The altitude to the hypotenuse of a right triangle (CB in this case) divides the hypotenuse into lengths such that the altitude is their geometric mean. That is ...

... CB = √(AC·CT) . . . . as above

This is true for any right triangle — another fact of geometry to put in your list of geometry facts.

3 0

2 years ago

The functions f(x) = −(x − 1)2 5 and g(x) = (x 2)2 − 3 have been rewritten using the completing-the-square method. apply your kn

ale4655 [162]

The vertex of the function f(x) exists (1, 5), the vertex of the function g(x) exists (-2, -3), and the vertex of the function f(x) exists maximum and the vertex of the function g(x) exists minimum.

<h3>How to determine the vertex for each function is a minimum or a maximum? </h3>

Given:

$\mathrm{f}(\mathrm{x})=-(\mathrm{x}-1)^{2}+5$ and

$\mathrm{g}(\mathrm{x})=(\mathrm{x}-2)^{2}-3$

The generalized equation of a parabola in the vertex form exists

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

Vertex of the function f(x) exists (1, 5).

Vertex of the function g(x) exists (-2, -3).

Now, if (a > 0) then the vertex of the function exists minimum, and if (a < 0) then the vertex of the function exists maximum.

The vertex of the function f(x) exists at a maximum and the vertex of the function g(x) exists at a minimum.

To learn more about the vertex of the function refer to:

brainly.com/question/11325676

#SPJ4

8 0

1 year ago

Explain the steps you would take to complete this conversion problem

OLEGan [10]

Answer:

20909.09 grams

Step-by-step explanation:

$\frac{46 \: lb}{1} \times \frac{1 \: kg}{2.2 \: lb} \times \frac{1000 \: g}{1 \: kg} = \\ \frac{46 \: lb}{2.2 \: lb} \times \frac{1 \: kg}{1} \: \times \frac{1000 \: g}{1 \: kg} = \\\frac{230 \: }{11 \:} \times \frac{1 \: kg}{1 \: kg \:}\times \frac{1000 \: g}{1 \:} = \\\frac{230 \: }{11 \:} \times \frac{1}{1}\times \frac{1000 \: g}{1 \:} = \\\ \frac{230}{11}g \times 1g \times \: 1000g$