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4 years ago

What is the area of a triangle whose vertices are J(−2, 1) , K(4, 3) , and L(−2, −5) ?

Mathematics

2 answers:

MArishka [77]4 years ago

8 0

Hello,

J (-2,1), K (4,3), and L (-2,-5)
Area of triangle: 18.00


REMEMBER: Notice that one side is vertical. Use that side as base.


Once again, Your answer is 18.

Hope this helps

Ksju [112]4 years ago

7 0

see attached picture:

area is around 18.96 units^2

round answer as needed, you don't say nearest whole number or nearest tenth

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From a box containing 10 cards numbered 1 to 10, four cards are drawn together. The probability that their sum is even is 21 21

ankoles [38]

Answer:

Step-by-step explanation:

We know that between 1 to 10 there are 5 even and 5 odd numbers.

We could get 4 even cards , 4 odd cards or 2 odd and 2 even cards

Let´s check all this combinations

Case 1: When all 4 numbers are even:

We are going to take 4 of the 5 even numbers in the box so we have

$5C4=5$

Case 2: When all 4 numbers are odd:

We are going to take 4 of the 5 odd numbers in the box, so we have

$5C4=5$

Case 3: When 2 are even and 2 are odd:

We are giong to take 2 from 5 even and odd cards in the box so we have

$5C2 * 5C2$

Remember that we obtain the probability from

$\frac{Number-of-favourable-Outcome}{Total-number-of-outcomes}$

So we have the number of favourable outcomes but we need the Total cases for drawing four cards, so we have that:

We are taking 4 of the 10 cards:

$10C_4=210$

Hence we have that the probability that their sum is even

$\frac{5+5+100}{210}=\frac{11}{21}$

8 0

3 years ago

A manufacturer produces bearings, but because of variability in the production process, not all of the bearings have the same di

Lena [83]

Answer:

Proportion of all bearings falls in the acceptable range = 0.9973 or 99.73% .

Step-by-step explanation:

We are given that the diameters have a normal distribution with a mean of 1.3 centimeters (cm) and a standard deviation of 0.01 cm i.e.;

Mean, $\mu$ = 1.3 cm and Standard deviation, $\sigma$ = 0.01 cm

Also, since distribution is normal;

Z = $\frac{X -\mu}{\sigma}$ ~ N(0,1)

Let X = range of diameters

So, P(1.27 < X < 1.33) = P(X < 1.33) - P(X <=1.27)

P(X < 1.33) = P( $\frac{X -\mu}{\sigma}$ < $\frac{1.33 -1.3}{0.01}$ ) = P(Z < 3) = 0.99865

P(X <= 1.27) = P( $\frac{X -\mu}{\sigma}$ < $\frac{1.27 -1.3}{0.01}$ ) = P(Z < -3) = 1 - P(Z < 3) = 1 - 0.99865

= 0.00135

P(1.27 < X < 1.33) = 0.99865 - 0.00135 = 0.9973 .

Therefore, proportion of all bearings that falls in this acceptable range is 99.73% .

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3 years ago

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strojnjashka [21]

Answer:

Step-by-step explanation:

f(x) = y = 0

so, f(x) = 0

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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