What is perpendicular to the line y= -1/4 x+3

True or false: y=35x is a linear equation.

adell [148]

Answer:

TRUE

Step-by-step explanation:

<u>what is a linear equation?</u>

It is said that a linear equation the equation which can be put in a form where there are variables, and there are coefficients, that are mainly and commonly 'actual numbers'

HOPE THIS HELPS!!

4 0

3 years ago

17. A bag contains 2x yellow counters, 4x + 6 red counters and 6x - 10 blue counters. a. Write an expression, in terms of x, for

Natalija [7]

Answer: 12x-4, 14 counters, 8 more counters

Step-by-step explanation: There are 2x yellow, 4x+6 red, and 6x-10 blue counters. We need to add all of this up to get the total number of counters. We first add the x's : 2x+4x+6x = 12x. Then we add the numbers. 6+ (-10) is -4. So, our expression is 12x-4. Next, we need to find out how many blue counters are in the bag. We know that there are 44 total counters and we need to find x because all the counters have x in them. 12x-4 = 44. We first add 4 to both sides to get 12x= 48 and x = 4. blue has 6x-10 counters so, blue has 24-10 = 14 counters. Red has 16+6 = 22 counters. 22-14 = 8 more counters

7 0

2 years ago

How to do transformations

photoshop1234 [79]

If it is a Reflection you use-
X-axis: (x,-y)
y-axis: (-x, y)
y=x: (y, x)
y=-x: (-y, -x)
Rotations:
90 degree clockwise/ 270 degree counterclockwise is (y, -x)
180 degree is (-x, -y)
270 degree clockwise/ 90 degree counterclockwise is (-y, x)
Dilations are when it is multiplied by a number. So lets say your operation is dilated by 2. It will be (2x, 2y).
Last but not least is Translations. This is when it is added or subtracted to your operation. This means it goes up or down and left or right on your graph. Left and right are X, up and down are Y. Lets say we want to go up 2 and left 3. It will be (x-3, y+2).

3 0

4 years ago

Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen

k0ka [10]

Let $a,b,c$ be the randomly selected lengths. Without loss of generality, suppose $a[tex]P(A + B \ge C) = P(A + B - C \ge 0)$

where $A,B,C$ are independent random variables with the same uniform distribution on [0, 1].

By their mutual independence, we have

$P(A=a,B=b,C=c) = P(A=a) \times P(B=b) \times P(C=c)$

so that the joint density function is

$P(A=a,B=b,C=c) = \begin{cases}1 & \text{if }(a,b,c)\in[0,1]^3 \\ 0 & \text{otherwise}\end{cases}$

where $[0,1]^3=[0,1]\times[0,1]\times[0,1]$ is the cube with vertices at (0, 0, 0) and (1, 1, 1).

Consider the plane

$a + b - c = 0$

with $(a,b,c)\in\Bbb R^3$ . This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)

The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region $a+b-c\ge0$ , since $0+0-1=-1$ . So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.

Then we have

$\displaystyle P(A+B-C\ge0) = 1 - P(A+B-C < 0) \\\\ ~~~~~~~~ = 1 - \int_0^1\int_0^{1-a}\int_{a+b}^1 P(A=a,B=b,C=c) \, dc\,db\,da \\\\ ~~~~~~~~ = 1 - \int_0^1 \int_0^{1-a} (1 - a - b) \, db \, da \\\\ ~~~~~~~~ = 1 - \int_0^1 \frac{(1-a)^2}2\,da \\\\ ~~~~~~~~ = 1 - \frac16 = \boxed{\frac56}$

5 0

2 years ago

Helpppppp <br> on a timer so need ASAP

leonid [27]

It would be 4 and 6 i solved it

7 0

3 years ago