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3 years ago

GEOMETRY PROOFS:

Mathematics

1 answer:

Marina CMI [18]3 years ago

8 0

Answer:

Statement 2: ∠BAC = ∠CZY

Reason 3: Vertically opposite angles

Statement 4: YC = CB

Reason 5: AAS (Angle, Angle, Side)

Reason 6: Corresponding sides, congruent triangles

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Please help me solve this HCF

Rina8888 [55]

Answer:

(x+y)

Step-by-step explanation:

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

Thus, the HCF is (x+y).

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2 years ago

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Find the domain and range of function f(x) is equal to root √x-1

Vladimir79 [104]

Answer:

Step-by-step explanation:

If you meant f(x) = √x - 1, then the domain is [0, infinity).

If you meant f(x) = √(x-1), then the domain is [1, infinity).

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3 years ago

What is the slope of y = -4?

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Answer:

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3 years ago

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18. At the movie theater, the 7:00 showing sold 75% of the tickets available.

attashe74 [19]

Answer:

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2 years ago

Find the missing measure for a right circular cone given the following information. Find V if r = 2 and L.A. = 12. (16/3)√2 16√2

aksik [14]

Volume of cone is given by formula

$V = \frac{\pi r^{2}h}{3}$ -------------------------------------------(1)

We are already given r =2, so we need to find h and then we can find volume

So missing meausre is height here ------------------------------------------------------------------------------

To find h we will use lateral area = $12\pi$ given

Formula for lateral area of cone is $\pi r\sqrt{r^{2}+ h^{2}}$

so we have $12 = \pi r\sqrt{r^{2}+ h^{2}}$

Now plug r as 2 in this equation

$12 \pi = \pi (2)\sqrt{2^{2}+ h^{2}}$

Now solve for h as shown and get h by itself

$\frac{12\pi}{2\pi} = \frac{2\pi\sqrt{4+h^{2}}}{2 \pi}$

$6 = \sqrt{4+h^{2}}$

$6^{2} = \sqrt{4+h^{2}} ^{2}$

$36} = 4+h^{2}$ $36 - 4 = 4+h^{2} -4$

$32 = h^{2}$ $\sqrt{32} = \sqrt{h^{2}}$

$\sqrt{16 \times 2} = h$ $\sqrt{16} \times \sqrt{2} = h$

$4\sqrt{2} = h$ ----------------------------------------------------(2)

Now plug 2 in r place and $4\sqrt{2}$ in h place in volume formula given in (1)

$V = \frac{\pi r^{2}h}{3}$

$V = \frac{\pi (2)^{2}4\sqrt{2}}{3} [/tex][tex] V = \frac{\pi (4)4\sqrt{2}}{3}$

$V = \frac{16 \pi \sqrt{2}}{3}$

so choice (1) is right answer

$(1) \frac{16 \sqrt{2}\pi}{3}$

correct answer --------------------------------------------------------------------------------------------------

$(2) 16 \sqrt{2} \pi$

incorrect answer as correct volume answer we got as $\frac{16 \sqrt{2}\pi}{3}$ --------------------------------------------------------------------------------------------------

$(3) 4 \sqrt{38} \pi$

incorrect answer as correct volume answer we got as $\frac{16 \sqrt{2}\pi}{3}$ --------------------------------------------------------------------------------------------------------

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3 years ago

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