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3 years ago

In this figure, what can be concluded about the distance from point B to CD?

Mathematics

2 answers:

maxonik [38]3 years ago

4 0

It is the same distance from the point A to CD

Ilia_Sergeevich [38]3 years ago

3 0

Answer:

Option D is right

Step-by-step explanation:

Given in the picture is a line segment AB.

From A and B an arc is drawn with equal radius to intersect at point C above AB and point D below AB.

CD is joined. Let E be the point of intersection of AB and CD

Now we find that CD is the perpendicular bisector of AB. (This is because AC=CB, AD=DB and CD is common. Hence ACD and ABD are congruent triangles by SSS theorem

By CPCT property for triangles, we get AE= EB and also

Angle AEC =BEC =90

Hence distance of A from CD is AE which equals BE, the distance of B from CD

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In most environments and cells themselves, there are

Salsk061 [2.6K]

Answer:

The cells membrane is selectively permeable, meaning only certain substances can easily pass through it. Only small, non polar molecules can easily pass through the small spaces between the hydrophobic heads and phospholipids. Water would have trouble due to it being a polar molecule. Proteins and carbs are quite large, making it difficult for them to pass through. Salt, which ionizes water, would also have difficulty passing through as it is a charged particle. Oxygen, on the other hand, is a non polar molecule that is small, so it can easily pass through the phospholipid bilayer.

Step-by-step explanation:

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3 years ago

melomori [17]

You multiply straight across, and one number is a fraction and the other is not.

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3 years ago

You roll two fair dice, one green and one red. (a) Are the outcomes on the dice independent? Yes No (b) Find P(1 on green die an

ASHA 777 [7]

Answer:

Yes

$P(1 \& 5) = \frac{1}{36}$

$P(5 \& 1) = \frac{1}{36}$

$P(5 \& 1 | 1\& 5 ) = \frac{1}{18}$

Step-by-step explanation:

From the question we are told that

Two fair dice, one green and one red were rolled

Generally the outcomes on the dice independent because the outcome on the first dice is not affected by the second die

Generally the probability of getting a 1 on a dice rolled is $P(1) = \frac{1}{6}$

the probability of getting a 5 on a dice rolled is $P(5) = \frac{1}{6}$

Generally the probability of P(1 on green die and 5 on red die) is mathematically represented as

$P(1 \& 5) = \frac{1}{6} *\frac{1}{6}$

$P(1 \& 5) = \frac{1}{36}$

Generally the probability of P(5 on green die and 1 on red die) is mathematically represented as

$P(5 \& 1) = \frac{1}{6} *\frac{1}{6}$

$P(5 \& 1) = \frac{1}{36}$

Generally the probability of P((1 on green die and 5 on red die) or (5 on green die and 1 on red die)) is mathematically represented as

$P(5 \& 1 | 1\& 5 ) = \frac{1}{36} + \frac{1}{36}$

$P(5 \& 1 | 1\& 5 ) = \frac{1}{18}$

8 0

3 years ago

Graph

posledela

Answer:

Step-by-step explanation:

3 0

3 years ago

last year 1/5 of the students in your class played a sport this year 10 students join the class five of the new students play a

tino4ka555 [31]

Answer:

Step-by-step explanation:

last year 1/5 of the class played sports, so s = (1/5) c

the next year, s increased by 5, c increased by 10 and the fraction changed to 1/4, so for the next year (s+5) = (1/4)(c+10)

To solve this we can substitute s = (1/5)c from the first equation into the second, so

((1/5)c + 5)=(1/4)(c+10), simplify both sides

(1/5)c + 5 = (1/4)c + 10/4, simplify 10/4 to 5/2

(1/5)c + 5 = (1/4)c + 5/2, multiply both sides by 20 to eliminate fractions (least common multiple of 2, 4 and 5)

4c + 100 = 5c + 50, subtract 4c, subtract 50 from both sides

50 = c, number of students in class last year was 50, this year is 10 more, so this year is 60

5 0

2 years ago