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Angelina_Jolie [31]

4 years ago

12

There are 10 to the 6th power millimeters in one kilometer. The distance from Dana’s house to her uncles house is 4 to the 4th p

ower kilometers. What is this distance in millimeters

1 answer:

Alex4 years ago

8 0

The answer is essentially 4^4 * 10^6

4^4 is 256, and 10^6 = 1,000,000

Knowing this, the answer should be 256,000,000

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Evaluate log6 base 10+ log45 base 10 - log27 base 10

Arlecino [84]

Answer:

The correct answer is B.

Explanation:

log106 + log1045 - log1027

log10

6

×

45

27

= log102 x 5

log1010 = 1

Step-by-step explanation:

4 0

3 years ago

110° 4tº + 90° t = degrees<br>need help please help me<br>

katen-ka-za [31]

Answer:

t = 5

Step-by-step explanation:

110-90 = 20

20/4 = 5

4 0

3 years ago

Write an equation of the line that passes through the given point and is parallel to the given line.

Margaret [11]

Answer:

Part 40) $y=-2x+9$

Part 41) $y=5x+6$

Part 42) $y=\frac{2}{3}x+\frac{4}{3}$

Step-by-step explanation:

Remember that

If two lines are parallel, then their slopes are the same

Part 40) Write an equation of the line that passes through the given point

and is parallel to the given line

we have

Point (4,1)

Given line y=-2x+7

step 1

Find the slope of the given line

The given line is a equation of the line in slope intercept form

$y=mx+b$

where

m is the slope

b is the y-intercept

so

The slope of the given line is

$m=-2$

step 2

Find the y-intercept b

we have

$m=-2$

$(4,1)$

substitute in the linear equation

$1=(-2)4+b$

solve for b

$b=1+8$

$b=9$

step 3

Find equation of the line that passes through the given point and is parallel to the given line

we have

$m=-2$

$b=9$

substitute

The equation of the line is

$y=-2x+9$

Part 41) Write an equation of the line that passes through the given point

and is parallel to the given line

we have

Point (0,6)

Given line y=5x-3

step 1

Find the slope of the given line

The given line is a equation of the line in slope intercept form

$y=mx+b$

where

m is the slope

b is the y-intercept

so

The slope of the given line is

$m=5$

step 2

Find the y-intercept b

$b=6$ ----> because the y-intercept is the point (0,6) (value of y when the value of x is equal to zero)

step 3

Find equation of the line that passes through the given point and is parallel to the given line

we have

$m=5$

$b=6$

substitute in the linear equation

$y=5x+6$

Part 42) Write an equation of the line that passes through the given point

and is parallel to the given line

we have

Point (-5,-2)

Given line y=(2/3)x+1

step 1

Find the slope of the given line

The given line is a equation of the line in slope intercept form

$y=mx+b$

where

m is the slope

b is the y-intercept

so

The slope of the given line is

$m=\frac{2}{3}$

step 2

Find the y-intercept b

we have

$m=\frac{2}{3}$

$(-5,-2)$

substitute in the linear equation

$-2=(\frac{2}{3})(-5)+b$

solve for b

$-2=-\frac{10}{3}+b$

$b=-2+\frac{10}{3}$

$b=\frac{4}{3}$

step 3

Find equation of the line that passes through the given point and is parallel to the given line

we have

$m=\frac{2}{3}$

$b=\frac{4}{3}$

substitute

The equation of the line is

$y=\frac{2}{3}x+\frac{4}{3}$

3 0

3 years ago

Line segment AB was dilated to create line segment A'B'

MrMuchimi

Answer:

13.6

Step-by-step explanation:

The measure of the segment AQ is related to the measure of the segment A'Q by the dilation factor as follows:

dilation factor = A'Q/AQ

AQ = A'Q/dilation factor

Replacing with data:

AQ = 2.4/0.15 = 16

Then, the distance between points A and A' is

x = 16 - 2.4 = 13.6

5 0

3 years ago

22: A 2.0 kg object moves along the x-axis under the influence of an unknown force. Its potential energy due to the force is U(x

aleksandr82 [10.1K]

Answer: a) -13.25 N, b) 1.78 m/sec

Step-by-step explanation:

Since we have given that

$U(x)=ax^4+bx^2$

where

$a=2.5\ J/m^4\\\\b=12\ J/m^2$

As we know that

$F=\dfrac{-dU}{dx}=-(4ax^3+2bx)$

Here, x= 0.5 m

So, it becomes,

$F=-(4\times 2.5\times (0.5)^3+2\times 12\times 0.5)=-13.25\ N$

Now, we need to find the velocity when it crosses the origin,

$F=mg\\\\F=m\dfrac{dv}{dt}\\\\F=mv\dfrac{dv}{dx}\\\\-(4ax^3+2bx)dx=mvdv\\\\\int\limits^{0.5}_0 {-(4ax^3+2bx)} \, dx =\int\limits {mv} \, dv\\\\ \dfrac{-4ax^4}{4}+\dfrac{2bx^2}{2}|_0^{0.5}=\dfrac{mv^2}{2}\\\\ax^4+bx^2|_0^{0.5}=\dfrac{mv^2}{2}\\\\2.5(0.5)^4+12(0.5)^2=\dfrac{2v^2}{2}\\\\0.15625+3=v^2\\\\3.15625=v^2\\\\\sqrt{3.15625}=v\\\\v=1.78$

Hence, a) -13.25 N, b) 1.78 m/sec

7 0

4 years ago