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IgorLugansk [536]

3 years ago

6

what is the equation in slope intercept form of the line that crosses the x-axis at -2 and is perpendicular to the line represen

ted by y equals negative 1/2 x + 3

1 answer:

Verdich [7]3 years ago

6 0

Answer:

y = -2x - 4

Step-by-step explanation:

So we have the x intercept and a line perpendicular to the one we are trying to find.

Perpendicular deals with the slope. Perpendicular slopes are negative reciprocals of each other, or in other words, if the slope is m the negative reciprocal is -1/m.

the slope of the perpendicular line is 1/2 so that means the negative reciprocal is -2.

x intercept is when y=0, so the point is (-2, 0). So that means the equation to make this point with the slope of -2 will look like this

y = mx + b

0 = -2*-2 + b

-4 = b

Now to get into the line we just plug in m and b into y = mx + b

y = mx + b

y = -2x - 4

And to check if you plug in -4 you'll get y = 0

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HELP ASAP What is the explicit formula for this sequence 5,2,-1,-4

Setler79 [48]

Answer:

For an nth term U(n) = a + (n - 1)d

n = number of terms

a = first term

d = common difference

a = 5

d = 2 - 5 = - 3

U(n) = 5 + (n -1)-3

= 5 - 3n + 3

= 8 - 3n

The formula for the sequence is 8 - 3n

Hope this helps

3 0

2 years ago

4-b/12=-3<br> What's b equal

NNADVOKAT [17]

Answer:

19-12=7, so 4-7=-3, so b is equal to 19

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2 years ago

Which of these relations on the set {0, 1, 2, 3} are equivalence relations? If not, please give reasons why. (In other words, if

Delicious77 [7]

Answer:

(1)Equivalence Relation

(2)Not Transitive, (0,3) is missing

(3)Equivalence Relation

(4)Not symmetric and Not Transitive, (2,1) is not in the set

Step-by-step explanation:

A set is said to be an equivalence relation if it satisfies the following conditions:

Reflexivity: If $\forall x \in A, x \rightarrow x$
Symmetry: $\forall x,y \in A, $if x \rightarrow y,$ then y \rightarrow x$
Transitivity: $\forall x,y,z \in A, $if x \rightarrow y,$ and y \rightarrow z, $ then x \rightarrow z$

(1) {(0,0), (1,1), (2,2), (3,3)}

(3) {(0,0), (1,1), (1,2), (2,1), (2,2), (3,3)}

The relations in 1 and 3 are Reflexive, Symmetric and Transitive. Therefore (1) and (3) are equivalence relation.

(2) {(0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)}

In (2), (0,2) and (2,3) are in the set but (0,3) is not in the set.

Therefore, It is not transitive.

As a result, the set (2) is not an equivalence relation.

(4) {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3)}

(1,2) is in the set but (2,1) is not in the set, therefore it is not symmetric

Also, (2,0) and (0,1) is in the set, but (2,1) is not, rendering the condition for transitivity invalid.

5 0

2 years ago

Pam has 90 m of fencing to enclose an area in a petting zoo with two dividers to separate three types of young animals. The thre

andrew-mc [135]

Answer:

The area function is

$A=\frac{135}{2}x-\frac{9}{2}x^2$ .

The domain and range of A is $(0,15m)$ and $(0, 253.125 m^2]$ .

Step-by-step explanation:

The given length of fencing is $90 m$ .

Let the length and width of each pen be $x$ and $y$ respectively as shown in the figure.

As there are 3 pens, so, the total area,

$A= 3 xy \;\cdots (i)$

From the figure the total length of fencing is $6x+4y$ .

Here, for a significant area for the animals, $x>0$ as well as $y>0$ as $x$ and $y$ are the sides of ben.

From the given value:

$6x+4y=90\;\cdots (ii)$

$\Rightarrow y=\frac {45}{2}-\frac{3x}{2}$

Now, from equation (i)

$A=3x\left(\frac {45}{2}-\frac{3x}{2}\right)$

$\Rightarrow A=\frac{135}{2}x-\frac{9}{2}x^2\;\cdots (iii)$

This is the required area function in the terms of variable $x$ .

For the domain of area function, from equation (ii)

$x=15-\frac{2y}{3}$

$\Rightarrow x$ [as y>0]

So, the domain of area function is $(0,15m)$ .

For the range of area function:

As $x \rightarrow 0$ or $y\rightarrow 0$ , then $A\rightarrow 0$ [from equation (i)]

$\Rightarrow A>0$

Now, differentiate the area function with respect to $x$ .

$\frac {dA}{dx}=\frac{135}{2}-9x$

Equate $\frac {dA}{dx}$ to zero to get the extremum point.

$\frac {dA}{dx}=0$

$\Rightarrow \frac{135}{2}-9x=0$

$\Rightarrow x=\frac{15}{2}$

Check this point by double differentiation

$\frac {d^2A}{dx^2}=-9$

As, $\frac {d^2A}{dx^2}$ , so, point $x=\frac{15}{2}$ is corresponding to maxima.

Put this value back to equation (iii) to get the maximum value of area function. We have

$A=\frac{135}{2}\times \frac {15}{2}-\frac{9}{2}\times \left(\frac {15}{2}\right)^2$

$\Rightarrow A=253.125 m^2$

Hence, the range of area function is $(0, 253.125 m^2]$ .

4 0

3 years ago

QUESTION NUMBER ONE

Mashutka [201]

Answer:

$68.85

Step-by-step explanation:

19% of $85 is $16.15

$85 - $16.15 = $68.85

5 0

2 years ago