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3 years ago

What is sin 78° rounded to the nearest hundredth

Mathematics

2 answers:

e-lub [12.9K]3 years ago

7 0

Use a calculator. The sine of 78 degrees is .98.

sergejj [24]3 years ago

5 0

I have to write 20 characters... 0.98, 0 being a whole, 9 being a tenth, 9 being a hundredth.

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3b^2-2c^5+23+a^2-6c^5-2b^2-4a^2-11

Komok [63]

Answer:

=3a^

Step-by-step explanation:

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3 years ago

Find h. 10 cm h = ✓ [?] cm 8 cm

vaieri [72.5K]

Answer:

√86 cm

Step-by-step explanation:

radius² + h² = slope² slope = 10 radius = 4

h² = slope² - radius²

h = √(slope² - radius²)

= √(10² - 4²)

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3 years ago

(5+3i)(4+2i) multiply

nika2105 [10]

Answer:

14+22i

Step-by-step explanation:

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3 years ago

Given AG bisects CD, IJ bisects CE, and BH bisects ED. Prove KE = FD.

OverLord2011 [107]

When a line is bisected, the line is divided into equal halves.

See below for the proof of $\mathbf{KE \cong FD}$

The given parameters are:

AC bisects CD
IJ bisects CE
BH bisects ED

By definition of segment bisection, we have:

$\mathbf{CK \cong KE}$
$\mathbf{EF \cong FD}$
$\mathbf{CE \cong ED}$

By definition of congruent segments, the above congruence equations become:

$\mathbf{CK = KE}$
$\mathbf{EF = FD}$
$\mathbf{CE = ED}$

By segment addition postulate, we have:

$\mathbf{CE = CK + KE}$
$\mathbf{ED = EF + FD}$

Substitute $\mathbf{ED = EF + FD}$ in $\mathbf{CE = ED}$

$\mathbf{CK + KE = EF + FD}$

Substitute $\mathbf{CK = KE}$ and $\mathbf{EF = FD}$

$\mathbf{KE + KE = FD + FD}$

Simplify

$\mathbf{2KE = 2FD}$

Apply division property of equality

$\mathbf{KE = FD}$

By definition of congruent segments

$\mathbf{KE \cong FD}$

Read more about proofs of congruent segments at:

brainly.com/question/11494126

6 0

3 years ago

Complete the equation describing how

Stolb23 [73]

Uh idek but i think the relationship is that x is plus 1 every time and y is plus 4 every time

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3 years ago