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3 years ago

Find the slope of the line through each pair of points (-12,0) (20,0)

Mathematics

1 answer:

AlexFokin [52]3 years ago

8 0

$\frac{y2}{x2} - \frac{y1}{x1}$
That is the formula to find the slope. Hope it will help you.

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Round 863 to the nearest ten

anygoal [31]

Step-by-step explanation:

$863 \approx \: 860$

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4 years ago

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A Kite of Area 255 m2 has one diagonal of length 9m . find the length of the other diagonal

Bas_tet [7]

$\\ \rm\longmapsto Area=255m^2$

$\\ \rm\longmapsto LB=255$

$\\ \rm\longmapsto 9L=255$

$\\ \rm\longmapsto L=\dfrac{255}{9}$

$\\ \rm\longmapsto L=28.3m$

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3 years ago

Felicia drew a triangle with side lengths that measured 15 cm, 18 cm and 27 cm. To determine if Felicia drew a right triangle, s

Charra [1.4K]

Answer:

The equation should be h²=P²+b²

27²=15²+18²

729=225+324

729=549

thus, it is not right angle triangle

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3 years ago

Find all ratios that are equivalent to 4 : 6 and have a first term that is less than 40.

kherson [118]

Answer:

2: 3, 4: 6, 6: 9, 8: 12 , 10: 15, 12: 18, 14: 21, 16: 24, 18: 27, 20: 30, 22: 33, 24:36, 26: 39, 28: 42, 30: 45, 32: 48, 34: 51, 36: 54, 38: 57

Step-by-step explanation:

The ratio 4: 6 can be simplified to 2:3 by dividing by 2 to each side of the ratio. After simplified to 2:3, you add the ration to itself to get the rest of the terms.

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3 years ago

Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%. Find

Papessa [141]

Answer:

1) $\text{P(at least one boy and one girl)}=\frac{3}{4}$

2) $\text{P(at least one boy and one girl)}=\frac{3}{8}$

3) $\text{P(at least two girls)}=\frac{1}{2}$

Step-by-step explanation:

Given : Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%.

To Find : The probability of each event.

1) P(at least one boy and one girl)

2) P(two boys and one girl)

3) P(at least two girls)

Solution :

Let's represent a boy with B and a girl with G

Mr. and Mrs. Romero are expecting triplets.

The possibility of having triplet is

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Total outcome = 8

$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}$

1) P(at least one boy and one girl)

Favorable outcome = BBG, BGB, BGG, GBB, GBG, GGB=6

$\text{P(at least one boy and one girl)}=\frac{6}{8}$

$\text{P(at least one boy and one girl)}=\frac{3}{4}$

2) P(at least one boy and one girl)

Favorable outcome = BBG, BGB, GBB=3

$\text{P(at least one boy and one girl)}=\frac{3}{8}$

3) P(at least two girls)

Favorable outcome = BGG, GBG, GGB, GGG=4

$\text{P(at least two girls)}=\frac{4}{8}$

$\text{P(at least two girls)}=\frac{1}{2}$

4 0

4 years ago

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