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densk [106]
3 years ago
9

Plz help! Pretty easy problems!

Mathematics
1 answer:
Julli [10]3 years ago
8 0

Answer:

The first is 9 sq.units

The second is 8 sq units

The third is 180mm squared

Step-by-step explanation:

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Evaluate start fraction 2 over 3 end fractionx for x = three-fourths
Umnica [9.8K]

Answer:

A

Step-by-step explanation:

7 0
3 years ago
Heya! ツ
fiasKO [112]

Answer:

See Below.

Step-by-step explanation:

In the given figure, O is the center of the circle. Two equal chords AB and CD intersect each other at E.

We want to prove that I) AE = CE and II) BE = DE

First, we will construct two triangles by constructing segments AD and CB. This is shown in Figure 1.

Recall that congruent chords have congruent arcs. Since chords AB ≅ CD, their respective arcs are also congruent:

\stackrel{\frown}{AB }\, \cong\, \stackrel{\frown}{CD}

Arc AB is the sum of Arcs AD and DB:

\stackrel{\frown}{AB}=\stackrel{\frown}{AD}+\stackrel{\frown}{DB}

Likewise, Arc CD is the sum of Arcs CB and DB. So:

\stackrel{\frown}{CD}=\stackrel{\frown}{CB}+\stackrel{\frown}{DB}

Since Arc AB ≅ Arc CD:

\stackrel{\frown}{AD}+\stackrel{\frown}{DB}=\stackrel{\frown}{CB}+\stackrel{\frown}{DB}

Solve:

\stackrel{\frown}{AD}\, \cong\,\stackrel{\frown}{CB}

The converse tells us that congruent arcs have congruent chords. Thus:

AD\cong CB

Note that both ∠ADC and ∠CBA intercept the same arc Arc AC. Therefore:

\angle ADC\cong \angle CBA

Additionally:

\angle AED\cong \angle CEB

Since they are vertical angles.

Thus:

\Delta AED\cong \Delta CEB

By AAS.

Then by CPCTC:

AE\cong CE\text{ and } BE\cong DE

5 0
3 years ago
Read 2 more answers
Drag the tiles to the boxes to form correct pairs. Polygon ABCD has sides with these lengths: , 5 units; , 4 units; , 4.5 units;
Airida [17]

Answer:

5 units

Step-by-step explanation:

5 0
3 years ago
two ferries start moving toward each other from opposite riverbanks, a and b. when they pass each other for the first time, the
Usimov [2.4K]

Let 'x' be the distance from THE far bank   where 700 is the distance to the NEAR bank

boat one has travelled 700    (rate = 700/unit time)         boat two has travelled   x    rate = x / unit time

boat one then travels   x  + 400  more      and boat two travels   700 + (700+x -400) more    when they meet

The time is the same      rate x time = distance      distance/rate = time     equate the distances divided by the respective rates

(700 + x + 400)/700     =    (  x  +  700     +  (700+x-400) )/x

1100x + x^2 = 1400x + 700000

x^2-300x -700000 = 0                         quadratic formula yields  x = 1000  

One boat travels 700   the other   1000 whe they first meet.....width of river =   700+ 1000 = 1700 m

3 0
3 years ago
What is the value of x?
blsea [12.9K]
Since these are supplementary angles  (two angles which sum to 180°) we can say:

4x+x=180  combine like terms on left side

5x=180  divide both sides by 5

x=36°
3 0
3 years ago
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