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3 years ago

What are the coordinates of the endpoints of the midsegment for △JKL that is parallel to KL¯¯¯¯¯?

Mathematics

2 answers:

goldenfox [79]3 years ago

8 0

Just took the quiz. The answer is (-1, -3) (1, -3)

Tom [10]3 years ago

6 0

The correct answer is (-1,-3) and (1,-3)

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PLEASE HELP ASAP 25pts

MrRissso [65]

The answer is B my friend

4 0

3 years ago

Read 2 more answers

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Ilia_Sergeevich [38]

Answer:

Then the solution is (4, 6).

Step-by-step explanation:

Let's use the substitution method:

First multiply the second equation by 4, obtaining 4y = -2x + 32.

Now substitute (-2x + 32) for 4y in the first equation:

3x + (-2x + 32) = 36, or

3x - 2x + 32 = 36. or

x = 4.

If x = 4, then the second equation yields y = (-1/2)(4) + 8, or

y = -2 + 8, or y = 6

Then the solution is (4, 6).

Check, using the first equation:

Does 3(4) + 4(6) = 36? Does 12 + 24 = 36? YES

7 0

2 years ago

What is the slope of the line represented by the equation y= 4/5x-3

guajiro [1.7K]

Answer:

$\frac{4}{5}$

Step-by-step explanation:

Slope of a line is the co efficient of x in the given line.

$y = \frac{4}{5} x - 3 \\ y = mx + b$

6 0

2 years ago

A small cylindrical has a diameter of 18mm and a height 12mm. Determine the minimum surface area of a box that can hold a row of

dimaraw [331]

Answer:

The surface are of a box = 1186.92 mm²

Step-by-step explanation:

Given in question as :

Diameter of cylinder = 18 mm

The height of cylinder = 12 mm

The surface area of cylinder = ( 2× $\pi$ ×r×h) + (2× $\pi$ ×r²)

where R is the radius , R = $\frac{Diameter}{2}$ = $\frac{18}{2}$

Or Radius r = 9 mm

Or, The surface area of cylinder= ( 2× $\pi$ ×9×12) + 2× $\pi$ ×(9)²)

Or , The surface area of cylinder=(678.24) + (508.68)

Or ,The surface area of cylinder= 1186.92 mm²

Hence the surface are of a box = 1186.92 mm² Answer

5 0

2 years ago

Suppose that X has an exponential distribution with mean equal to 10. Determine the following: a. P(X > 10) b. P(X > 20) c

GrogVix [38]

Answer:

(a) The value of P (X > 10) is 0.3679.

(b) The value of P (X > 20) is 0.1353.

(d) The value of x is 30.

Step-by-step explanation:

The probability density function of an exponential distribution is:

$f(x)=\lambda e^{-\lambda x};\ x>0, \lambda>0$

The value of E (X) is 10.

The parameter λ is:

$\lambda=\frac{1}{E(X)}=\frac{1}{10}=0.10$

(a)

Compute the value of P (X > 10) as follows:

$P(X>10)=\int\limits^{\infty}_{10} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{10} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{10}\\=|e^{-0.10 x} |^{\infty}_{10}\\=e^{-0.10\times10}\\=0.3679$

Thus, the value of P (X > 10) is 0.3679.

(b)

Compute the value of P (X > 20) as follows:

$P(X>20)=\int\limits^{\infty}_{20} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{20} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{20}\\=|e^{-0.10 x} |^{\infty}_{20}\\=e^{-0.10\times20}\\=0.1353$