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Vitek1552 [10]
3 years ago
9

How many orders of magnitude bigger is 23,000 than 56?

Mathematics
1 answer:
Daniel [21]3 years ago
4 0

Orders of Magnitude are in terms of scientific notation.

23,000 written in scientific notation is 2.3 x 10^4, it's magnitude is 4 ( the number 10 is raised to)

56 in scientific notation would be 5.6 x 10^1, its magnitude is 1.

4 - 1 = 3

2300 is 3 order of magnitudes larger.

The answer is 3.

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Evaluate $\frac{20}{101}-\frac{20}{99}$. Express your answer in simplest form.
m_a_m_a [10]
Detail: 37









Answer: 69


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Blick: 55
7 0
2 years ago
Which correctly describes how the graph of the inequality 6y − 3x > 9 is shaded? -Above the solid line
baherus [9]

The statement which correctly describes the shaded region for the inequality is \fbox{\begin\\\ Above the dashed line\\\end{minispace}}

Further explanation:

In the question it is given that the inequality is 6y-3x>9.  

The equation corresponding to the inequality 6y-3x>9 is 6y-3x=9.

The equation 6y-3x=9 represents a line and the inequality 6y-3x>9 represents the region which lies either above or below the line 6y-3x=9.

Transform the equation 6y-3x=9 in its slope intercept form as y=mx+c, where m represents the slope of the line and c represents the y-intercept.  

y-intercept is the point at which the line intersects the y-axis.  

In order to convert the equation 6y-3x=9 in its slope intercept form add 3x to equation 6y-3x=9.  

6y-3x+3x=9+3x

6y=9+3x

Now, divide the above equation by 6.  

\fbox{\begin\\\math{y=\dfrac{x}{2}+\dfrac{1}{2}}\\\end{minispace}}

Compare the above final equation with the general form of the slope intercept form \fbox{\begin\\\math{y=mx+c}\\\end{minispace}}.  

It is observed that the value of m is \dfrac{1}{2} and the value of c is \dfrac{3}{2}.

This implies that the y-intercept of the line is \dfrac{3}{2} so, it can be said that the line passes through the point \fbox{\begin\\\ \left(0,\dfrac{3}{2}\right)\\\end{minispace}}.

To draw a line we require at least two points through which the line passes so, in order to obtain the other point substitute 0 for y in 6y=9+3x.  

0=9+3x

3x=-9

\fbox{\begin\\\math{x=-3}\\\end{minispace}}  

This implies that the line passes through the point \fbox{\begin\\\ (-3,0)\\\end{minispace}}.  

Now plot the points (-3,0) and \left(0,\dfrac{3}{2}\right) in the Cartesian plane and join the points to obtain the graph of the line 6y-3x=9.  

Figure 1 shows the graph of the equation 6y-3x=9.

Now to obtain the region of the inequality 6y-3x>9 consider any point which lies below the line 6y-3x=9.  

Consider (0,0) to check if it satisfies the inequality 6y-3x>9.  

Substitute x=0 and y=0 in 6y-3x>9.  

(6\times0)-(3\times0)>9  

0>9

The above result obtain is not true as 0 is not greater than 9 so, the point (0,0) does not satisfies the inequality 6y-3x>9.  

Now consider (-2,2) to check if it satisfies the inequality 6y-3x>9.  

Substitute x=-2 and y=2 in the inequality 6y-3x>9.  

(6\times2)-(3\times(-2))>9  

12+6>9  

18>9  

The result obtain is true as 18 is greater than 9 so, the point (-2,2) satisfies the inequality 6y-3x>9.  

The point (-2,2) lies above the line so, the region for the inequality 6y-3x>9 is the region above the line 6y-3x=9.  

The region the for the inequality 6y-3x>9 does not include the points on the line 6y-3x=9 because in the given inequality the inequality sign used is >.

Figure 2 shows the region for the inequality \fbox{\begin\\\math{6y-3x>9}\\\end{minispace}}.

Therefore, the statement which correctly describes the shaded region for the inequality is \fbox{\begin\\\ Above the dashed line\\\end{minispace}}

Learn more:  

  1. A problem to determine the range of a function brainly.com/question/3852778
  2. A problem to determine the vertex of a curve brainly.com/question/1286775
  3. A problem to convert degree into radians brainly.com/question/3161884

Answer details:

Grade: High school

Subject: Mathematics  

Chapter: Linear inequality

Keywords: Linear, equality, inequality, linear inequality, region, shaded region, common region, above the dashed line, graph, graph of inequality, slope, intercepts, y-intercept, 6y-3x=9, 6y-3x>9, slope intercept form.

4 0
3 years ago
Read 2 more answers
A LOT OF POINTS pls help and answer correctly if not I will report ty asap !
Solnce55 [7]

Answer:

hey

Step-by-step explanation:

to calculate perimeter, you require length,breadth

since points are given,find distance between points using distance formulae

<h2>root over (x2-x1)² +(y2-y1)²</h2>

______________________

<h3>Point H =(6,7)</h3><h3>Point I=(-6,-9)</h3><h3>Point J=(-10,-6)</h3><h3>Point G=2,10)</h3>

Now find distance between HI and IJ

HI gives the distance - length of rectangle

IJ gives the distance - breadth of rectangle

use the formulae ,2(length+breadth) to get perimeter

______________________

HI= root over 144+256= root 400

HI=20

IJ= root over 16+9= root 25

IJ=5

PERIMETER =2(20+5)

2×25

=50

6 0
3 years ago
Please show me step by step on how to solve this.
Readme [11.4K]
Well, you have to be given an interval. That is not an interval.

You should be given something like 1 \leq x \leq 2

If you add those fractions up you'll just get a point on the number line.

Find the LCD for all of them which is 9/6 - 4/6 + 3/6 = 8/6 = 2/3

You'll end up with a point on 2/3 
7 0
3 years ago
The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f
Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

h(t)=-16t^{2}+25t+5

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

so

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

h(t)=-16(t^{2}-\frac{25}{16}t)+5

Complete the square

h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}

h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}

Rewrite as perfect squares

h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}

The vertex is the point (\frac{25}{32},\frac{945}{64})

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}

16(t-\frac{25}{32})^{2}=\frac{945}{64}

(t-\frac{25}{32})^{2}=\frac{945}{1,024}

square root both sides

(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}

t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}

the positive value is

t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec

8 0
3 years ago
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