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2 years ago

What type of association is shown in the scatter plot?

Mathematics

1 answer:

choli [55]2 years ago

8 0

Answer:

Step-by-step explanation:

It's going up, so its positive and you could draw a line straight through the points, because they are all very close .

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What is the solution of StartRoot x minus 4 EndRoot + 5 = 2?

Vlad1618 [11]

Answer:

Step-by-step explanation:

$\sqrt{x-4}+5=2 \\\\\sqrt{x-4}=-3 \\\\x-4=9 \\\\x=13$

Hope this helps!

6 0

2 years ago

Read 2 more answers

g My Notes Water flows from the bottom of a storage tank at a rate of r(t) = 200 − 4t liters per minute, where 0 ≤ t ≤ 50. Find

Arada [10]

Answer:

4,200 liters

Step-by-step explanation:

The flow rate is given by:

$r(t) = 200 - 4t$

Integrating the flow rate expression from t=0 to t=30 minutes, yields the total volume that flows out of the tank during that period:

$\int\limits^{30}_0 {r(t)} = \int\limits^{30}_0 {(200 - 4t} )\, dt \\V=(200t - 2t^2)|_0^{30}\\V= (200*30 -2*30^2)-(200*0 -2*0^2)\\V=4,200\ liters$

4,200 liters of water flow from the tank during the first 30 minutes.

3 0

2 years ago

Which of the following is the graph of the quadratic function? y=x^2+4-12

tiny-mole [99]

Answer:

Graph B

Step-by-step explanation:

The solutions to a quadratic equation are the points on which we have the graph of the curve touching the x-axis

Now the first thing we will do here is to solve the quadratic equation graph;

x^2 + 4x -12 = 0

x^2 +6x - 2x -12 = 0

x (x + 6) -2(x + 6) = 0

(x-2)(x + 6) = 0

x = 2 or -6

So the graph that touches the x-axis at the points x = 2 and x = -6 is the solution to the quadratic equation

Graph B is the closest to what we have as answer

5 0

3 years ago

Suppose u and v are functions of x that are differentiable at x=0 and that u(0)=3, u'(0)= -4, v(0)= 2, v'(0)= -7. Find the value

3241004551 [841]

$\bf \cfrac{d}{dx}[u(x)\cdot v(x)]\implies \cfrac{du}{dx}\cdot v(x)+u(x)\cdot \cfrac{dv}{dx}\quad 
\begin{cases}
u(0)=3\qquad u'(0)=-4
\\\\
v(0)=2\qquad v'(0)=-7
\end{cases}
\\\\
\left[ \cfrac{du}{dx} \right]_{x=0}\cdot v(0)+u(0)\cdot \left[ \cfrac{dv}{dx} \right]_{x=0}$

so.. hmm pretty sure you know what that is

7 0

2 years ago

Read 2 more answers

PLEASE HELP!! I do not understand this. And can you explain it for me when I use it again?

damaskus [11]

Let's start b writing down coordinates of all points:
A(0,0,0)
B(0,5,0)
C(3,5,0)
D(3,0,0)
E(3,0,4)
F(0,0,4)
G(0,5,4)
H(3,5,4)

a.) When we reflect over xz plane x and z coordinates stay same, y coordinate changes to same numerical value but opposite sign. Moving front-back is moving over x-axis, moving left-right is moving over y-axis, moving up-down is moving over z-axis.

A(0,0,0)
Reflecting
A(0,0,0)

B(0,5,0)
Reflecting
B(0,-5,0)

C(3,5,0)
Reflecting
C(3,-5,0)

D(3,0,0)
Reflecting
D(3,0,0)

b.)
A(0,0,0)
Moving
A(-2,-3,1)

B(0,-5,0)
Moving
B(-2,-8,1)

C(3,-5,0)
Moving
C(1,-8,1)

D(3,0,0)
Moving
D(1,-3,1)

7 0

2 years ago