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4 years ago

Enter the number in standard notation. 8.026 x 10-6 = What is the answer?

Mathematics

1 answer:

juin [17]4 years ago

4 0

Answer:

.000008026

Step-by-step explanation:

move the decimal forward 6 places

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in 1985, there were 285 cell phone subscribers in the small town of glenwood. the number of subscribers increased by 75% per yea

-Dominant- [34]

The calculations come out to be <span>2003.44921875, so there would be 2003 subscribers in 1990</span>

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3 years ago

Given A is between Y and Z and YA = 2x, AZ = 16x, and YZ = 4x+56, find AZ.<br> Plsss help

cupoosta [38]

Answer:

Step-by-step explanation:

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3 years ago

Given the triangles below what congruence theorem would prove that the two triangles were congruent.

BaLLatris [955]

The correct answer and im sure of it is 69

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4 years ago

The equation of a circle is x2 + y2 + 6x + 4y + 10 = 1. What is this equation written in its standard form?

Marrrta [24]

Option A: $(x+3)^2+(y+2)^2=4$ is the equation of the circle

Explanation:

The equation of the circle is $x^{2} +y^2+6x+4y+10=1$

We need to write the equation of the circle in standard form.

Let us subtract 10 from both sides of the equation.

Thus, we have,

$x^{2} +y^2+6x+4y=-9$

Grouping the terms,we have,

$(x^{2} +6x)+(y^2+4y)=-9$

Now, we shall complete the square, let us add and subtract the equation by 9 to write the term $\left(x^{2}+6 x\right)$ in the form of $(a+b)^2$

Thus, we have,

$(x^{2} +6x+9-9)+(y^2+4y)=-9$

Simplifying, we get,

$(x^{2} +6x+9)+(y^2+4y)=-9+9$

$(x+3)^2+(y^2+4y)=0$

Similarly, we shall complete the square for the term $\left(y^{2}+4 y\right)$ in the form of $(a+b)^2$

Thus, we have,

$(x+3)^2+(y^2+4y+4-4)=0$

Simplifying, we get,

$(x+3)^2+(y^2+4y+4)=4$

$(x+3)^2+(y+2)^2=4$

Thus, the equation of the circle in standard form is $(x+3)^2+(y+2)^2=4$

Hence, Option A is the correct answer.

3 0

4 years ago

20 25 30 30 31 40 41 49

Ludmilka [50]

I can see this is a pattern.. 5,5,0,1,1,5 etc.
Did you want us to tell you the rest or something?

3 0

3 years ago