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Kitty [74]
2 years ago
10

Consider the two triangles. Triangles A B C and H G I are shown. Angles A C B and H I G are right angles. The length of side A C

is 15 and the length of side C B is 20. The length of side H I is 12 and the length of I G is 9. To prove that the triangles are similar by the SAS similarity theorem, it needs to be shown that

Mathematics
2 answers:
zheka24 [161]2 years ago
8 0

Answer:

Both are similar by SAS similarity.

This SAS similarity is equivalent to the congruence.

Step-by-step explanation:

Step 1:

To prove that ACB and HIG as similar triangles.

We have to look upon the corresponding sides.

SAS= Side angle sides , there the angle must be in between two sides.

\angle ACB = \angle HIG

Lets work on the corresponding sides.

IG/AC = IH/AC

\frac{9}{15} = \frac{12}{20}

Reducing each to lowest form, we divide numerator and denominator by 3 for the 1st fraction and by 4 for the 2nd fraction.

We have

\frac{3}{5} = \frac{3}{5}

Both sides are equal.

So its proved that both are similar with SAS similarity theorem.

garri49 [273]2 years ago
7 0

Answer: C

Step-by-step explanation: I worked it out for you :)

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On the first day of a family road trip, Kai's family travels 365 miles.
Gekata [30.6K]
They traveled 292 miles on day two.

Known: On the first day they traveled 365 and on the second they traveled 20% less.

Solution:

If they traveled 20% less on the second day, that means they traveled 80% of the distance they traveled the first day.

365 miles * .8 = 292.

You could also solve this as:
20% of 365 is 73 miles
365 * .2 = 73.
So they traveled 73 less miles on the second day.
365 miles on the first day - 73 miles less on the second day = 292 miles.

I hope this helps!
7 0
1 year ago
Find FL if H is the circumcenter of EFG,
Zanzabum

The artistic crop isn't helpful; it cuts off some vertex names.

The circumcenter H is the meet of the perpendicular bisectors of the sides, helpfully drawn.  We have right triangle ELH, right angle L, so

EH² = HL² + EL²

EL = √(EH² - HL²) = √(5.06²-2.74²) ≈ 4.2539393507665339

Since HL is a perpendicular bisector of EF, we get congruent segments FL=EL.

Answer: 4.25

8 0
2 years ago
Complete the steps to find all zeroes of the function f(x) = x4 − 4x3 − 4x2 + 36x − 45. This function has roots.
xxTIMURxx [149]
Steps?

A graph shows zeros to be ±3. Factoring those out leaves the quadratic
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2 years ago
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If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe
Natalka [10]

Answer:

They encircle the planet 3.76\times 10^{11} times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is 6.022\times 10^{23}

Thus, the number of dust particles is:

2.5\times 6.022\times 10^{23}=15.055\times 10^{23}

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: 10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m

If we line up the particles the distance they could cover is:

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Circumference in meters:

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Therefore,

\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}

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8 0
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Answer:

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