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3 years ago

I don’t get this please help ASAP

Mathematics

1 answer:

jek_recluse [69]3 years ago

3 0

Answer:

AE = 70

Step-by-step explanation:

Given that ACE is a triangle and B , D , F are mid-points of AC, CE, AE respectively.

⇒ BF, FD, BD are mid-segments.

Mid-segment is a line segment joining mid points of two sides of a triangle,

And it as two properties :

1) mid-segment is always parallel to the third side

2) mid-segment is half in length of the third side

⇒ Here, BD is the mid-segment and is parallel to third side AE

And also BD is half of AE

⇒ AE = 2×BD = 2×35

⇒ AE = 70.

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3 years ago

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stealth61 [152]

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3 years ago

Read 2 more answers

Sara has 20 sweets. She has 12 liquorice sweets, 5 mint sweets and 3 humbugs. Sarah is going to take, at random, two sweets Work

Debora [2.8K]

Answer:

111 / 190

Step-by-step explanation:

Let us first compute the probability of picking 2 of each sweet. Take liquorice as the first example. There are 12 / 20 liquorice now, but after picking 1 there will be 11 / 19 left. Thus the probability of getting two liquorice is demonstrated below;

$12 / 20 * 11 / 19 = \frac{33}{95},\\Probability of Drawing 2 Liquorice = \frac{33}{95}$

Apply this same concept to each of the other sweets;

$5 / 20 * 4 / 19 = \frac{1}{19},\\Probability of Drawing 2 Mint Sweets = 1 / 19\\\\3 / 20 * 2 / 19 = \frac{3}{190},\\Probability of Drawing 2 Humbugs = 3 / 190$

Now add these probabilities together to work out the probability of drawing 2 of the same sweets, and subtract this from 1 to get the probability of not drawing 2 of the same sweets;

$33 / 95 + 1 / 19 + 3 / 190 = \frac{79}{190},\\1 - \frac{79}{190} = \frac{111}{190}\\\\$

The probability that the two sweets will not be the same type of sweet =

111 / 190

3 0

3 years ago

Anybody have a real answer??

Elena-2011 [213]

I think Its 66 cm. Hope This Helped! Good Luck...

5 0

3 years ago

find the measure of the side of the right triangle whose length is designated by a lowercase letter. Round answers to the neares

mario62 [17]

Ah yes, math, I believe the answer you are looking for, my young scholar, is A

8 0

3 years ago