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2 years ago

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ira [324]2 years ago

4 0

Answer:

10...

Step-by-step explanation:

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Help please which statement is true about water molecules? question 1 options: the oxygen atom has a negative charge. the hydrog

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Oxygen having a negative charge I believe

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3 years ago

Help me with this question I don’t really understand it:(

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Step-by-step explanation:

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2 years ago

NO LINKS!!!! Find the equation of the circle below

inysia [295]

Answer:

$(x+3)^2+(y-2)^2=9$

Step-by-step explanation:

<u>Equation of a circle</u>

$(x-a)^2+(y-b)^2=r^2$

where:

(a, b) is the center
r is the radius

From inspection of the diagram, the center of the circle <em>appears</em> to be at point (-3, 2), although this is not very clear. Therefore, a = -3 and b = 2.

Substitute these values into the general form of the equation of a circle:

$\implies (x-(-3))^2+(y-2)^2=r^2$

$\implies (x+3)^2+(y-2)^2=r^2$

Again, from inspection of the diagram, the <u>maximum vertical point</u> of the circle appears to be at y = 5. Therefore, to calculate the radius, subtract the y-value of the center point from the y-value of the maximum vertical point:

⇒ radius (r) = 5 - 2 = 3

Substitute the found value of r into the equation:

$\implies (x+3)^2+(y-2)^2=3^2$

Therefore, the final equation of the given circle is:

$\implies (x+3)^2+(y-2)^2=9$

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1 year ago

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Yesterday, Linda had 71 baseball cards. Today, she gave d away. Using d, write an expression for the number of cards Linda has l

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Answer:

71-d

Step-by-step explanation:

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2 years ago

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The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon

Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by $x=3.24t-4.20t^2+1.07t^3$ where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(1)=3.24(1)-4.20(1)^2+1.07(1)^3$

$x(1)=3.24-4.20+1.07$

$x(1)=0.11$

b) At t= 2 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(2)=3.24(2)-4.20(2)^2+1.07(2)^3$

$x(2)=6.48-16.8+8.56$

$x(2)=-1.76$

c) At t= 3 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(3)=3.24(3)-4.20(3)^2+1.07(3)^3$

$x(3)=9.72-37.8+28.89$

$x(3)=0.81$

d) At t= 4 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(4)=3.24(4)-4.20(4)^2+1.07(4)^3$

$x(4)=12.96-67.2+68.48$

$x(4)=14.24$

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

$\triangle x=x(4)-x(0)$

$\triangle x=14.24-0$

$\triangle x=14.24$

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity is given by,

$\triangle x=x(4)-x(2)$

$\triangle x=14.24-(-1.76)$

$\triangle x=14.24+1.76$

$\triangle x=16$

4 0

2 years ago