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2 years ago

Which of the following are equations for the line shown below? Check all that

Mathematics

1 answer:

Maurinko [17]2 years ago

4 0

ANSWER: A & B

FOR COORDINATES (2,2)
A) 2-2=-0.25(2-2)
0=0
B) 2-3=-0.25(2+2)
-1=-1

FOR COORDINATES (-2,3)
A) 3-2=-0.25(-2-2)
1=1
B) 3-3=-0.25(-2+2)
0=0

WHY ITS NOT C and D

COORDINATES (2,2)
C) 2+3=-0.25(2-2)
5=0 ZERO DOES NOT EQUAL 5
D) 2-3=-0.25(2-2)
-1=0 ZERO DOES NOT EQUAL -1

COORDINATES (-2,3)
C) 3+3=-0.25(-2-2)
6=1 SIX DOES NOT EQUAL 1
D) 3-3=-0.25(-2-2)
0=1 ZERO DOES NOT EQUAL 1

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Please help need answer ASAP

mariarad [96]

X = 71 degrees
And here is a sentence to fill the minimum word rule

3 0

3 years ago

C(25,10)<br> Find permutation and combination

Verizon [17]

C(25,10)= 3268760

P(25,10)=11861676288000

hope it helps but these are very big numbers :d

6 0

3 years ago

Louise begins factoring the polynomial, which has four terms.

ad-work [718]

Louise wants to factorize completely the given polynomial $4x^{3}+12x^{2}+5x+15$

Grouping first, second terms together and third, fourth terms together

= $4x^{3}+12x^{2}+5x+15$

Taking $4x^{2}$ common from first and second terms of the given expression and taking 5 common from the third and fourth terms.

= $=4x^{2}(x+3)+5(x+3)$

Taking (x+3) common from the given expression,

= $=(4x^{2}+5)(x+3)$ is the completely factored form.

5 0

3 years ago

Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

dexar [7]

Answer:

The value of the constant C is 0.01 .

Step-by-step explanation:

Given:

Suppose X, Y, and Z are random variables with the joint density function,

$f(x,y,z) = \left \{ {{Ce^{-(0.5x + 0.2y + 0.1z)}; x,y,z\geq0 } \atop {0}; Otherwise} \right.$

The value of constant C can be obtained as:

$\int_x( {\int_y( {\int_z {f(x,y,z)} \, dz }) \, dy }) \, dx = 1$

$\int\limits^\infty_0 ({\int\limits^\infty_0 ({\int\limits^\infty_0 {Ce^{-(0.5x + 0.2y + 0.1z)} } \, dz }) \, dy } )\, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y }(\int\limits^\infty_0 {e^{-0.1z} } \, dz }) \, dy }) \, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0{e^{-0.2y}([\frac{-e^{-0.1z} }{0.1} ]\limits^\infty__0 }) \, dy }) \, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}([\frac{-e^{-0.1(\infty)} }{0.1}+\frac{e^{-0.1(0)} }{0.1} ]) } \, dy }) \, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}[0+\frac{1}{0.1}] } \, dy }) \, dx =1$

$10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2y} }{0.2}]^\infty__0 }) \, dx = 1$

$10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2(\infty)} }{0.2}+\frac{e^{-0.2(0)} }{0.2}] } \, dx = 1$

$10C\int\limits^\infty_0 {e^{-0.5x}[0+\frac{1}{0.2}] } \, dx = 1$

$50C([\frac{-e^{-0.5x} }{0.5}]^\infty__0}) = 1$

$50C[\frac{-e^{-0.5(\infty)} }{0.5} + \frac{-0.5(0)}{0.5}] =1$

$50C[0+\frac{1}{0.5} ] =1$

$100C = 1$ ⇒ $C = \frac{1}{100}$

C = 0.01

3 0

3 years ago

Does -20 belong in whole numbers

AURORKA [14]

Oh Yes It Does Yes It dose

7 0

3 years ago