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allsm [11]
3 years ago
11

If a graph shows the function y=3x what are the coordinates of the intercepts

Mathematics
1 answer:
vladimir2022 [97]3 years ago
3 0

Answer:

graph{3x+5 [-10, 10, -5, 5]}

x

intercept:  

x

=

−

5

3

 

y

intercept:  

y

=

5

Explanation:

For a linear graph, the quickest way to sketch the function is to determine the  

x

and  

y

intercepts and draw a line between the two: this line is our graph.

Let's calculate the  

y

intercept first:

With any function,  

y

intercepts where  

x

=

0

.

Therefore, substituting  

x

=

0

into the equation, we get:

y

=

3

⋅

0

+

5

y

=

5

Therefore, the  

y

intercept cuts through the point (0,5)

Let's calculate the  

x

intercept next:

Recall that with any function:  

y

intercepts where  

x

=

0

.

The opposite is also true: with any function  

x

intercepts where  

y

=

0

.

If we substitute  

y

=

0

, we get:

0

=

3

x

+

5

Let's now rearrange and solve for  

x

to calculate the  

x

intercept.

−

5

=

3

x

 

−

5

3

=

x

Therefore, the  

x

intercept cuts through the point  

(

−

5

3

,

0

)

.

Now we have both the  

x

and  

y

intercepts, all we have to do is essentially plot both intercepts on a set of axis and draw a line between them

The graph of the function  

y

=

3

x

+

5

:

graph{3x+5 [-10, 10, -5, 5]}



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What is the numb bur 731,934,000 in standard form
vesna_86 [32]
That number is already in standard form.

If you need to write it in scientific notation, then do this.

<span>731,934,000

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The original decimal point was after the third zero, so now using the original number, count the number of decimal places you moved the decimal point. The decimal point moved 8 places to the left. That means you need an exponent of 8.

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8 0
4 years ago
The altitude of a triangle is a perpendicular segment from the vertex to a line containing the
azamat

Answer:

H=\sqrt{91}

Step-by-step explanation:

The altitude falls in the middle of the line.

So, the point at which it falls in the bottom line is the midpoint, so the length of the blue triangle would be 6/2 = 3

So, the blue triangle has length of 3 and height of H and hypotenuse of 10.

Now, we will use Pythagorean Theorem to solve for H. The theorem is:

Leg^2 + Another Leg^2 = Hypotenuse^2

We know one leg of the triangle is 3 and another leg is H and hypotenuse is 10, so we substitute and solve:

3^2 + H^2 = 10^2\\9+H^2=100\\H^2=100-9\\H^2=91\\H=\sqrt{91}

This is the exact answer for H

6 0
3 years ago
HELP ITS A TEST!!! IF YOU KNOW HOW TO DO THIS HELP PLEASE!
Naddik [55]

Answer:

Step-by-step explanation:

A = base *h /2

A = (4*15/2 )/2  = 15 in²

A = (1 and 3/7 *14/5 )/2  = (10/7 * 14/5 )/2 = 2 in²

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6 0
3 years ago
Construct a frequency distribution and a relative frequency distribution for the light bulb data with a class width of 20, start
k0ka [10]

Answer:

Step-by-step explanation:

Hello!

You have the information about light bulbs (i believe is their lifespan in hours) And need to organize the information in a frequency table.

The first table will be with a class width of 20, starting with 800. This means that you have to organize all possible observations of X(lifespan of light bulbs) in a class interval with an amplitude of 20hs and then organize the information noting their absolute frequencies.

Example

1) [800;820) only one observation classifies for this interval x= 819, so f1: 1

2)[820; 840) only one observation classifies for this interval x= 836, so f2: 1

3)[840;860) no observations are included in this interval, so f3=0

etc... (see attachment)

[ means that the interval is closed and starts with that number

) means that the interval is open, the number is not included in it.

fi: absolute frequency

hi= fi/n: relative frequency

To graph the histogram you have to create the classmark for each interval:

x'= (Upper bond + Lower bond)/2

As you can see in the table, there are several intervals with no observed frequency, this distribution is not uniform least to say symmetric.

To check the symmetry of the distribution is it best to obtain the values of the mode, median and mean.

To see if this frequency distribution has one or more modes you have to identify the max absolute frequency and see how many intervals have it.

In this case, the maximal absolute frequency is fi=6 and only one interval has it [1000;1020)

Mo= LB + Ai (\frac{D_1}{D_1+D_2} )\\

LB= Lower bond of the modal interval

D₁= fmax - fi of the previous interval

D₂= fmax - fi of the following interval

Ai= amplitude of the modal interval

Mo= 1000 + 20*(\frac{(6-3)}{(6-3)+(6-4)} )=1012

This distribution is unimodal (Mo= 1012)

The Median for this frequency:

Position of the median= n/2 = 40/2= 20

The median is the 20th fi, using this information, the interval that contains the median is [1000;1020)

Me= LB + Ai*[\frac{PosMe - F_{i-1}}{f_i} ]

LB= Lower bond of the interval of the median

Ai= amplitude of the interval

F(i-1)= acumulated absolute frequency until the previous interval

fi= absolute frequency of the interval

Me= 1000+ 20*[\frac{20-16}{6} ]= 1013.33

Mean for a frequency distribution:

X[bar]= \frac{sum x'*fi}{n}

∑x'*fi= summatory of each class mark by the frequency of it's interval.

∑x'*fi= (810*1)+(230*1)+(870*0)+(890*2)+(910*4)+(930*0)+(950*4)+(970*1)+(990*3)+(1010*6)+(1030*4)+(1050*0)+(1070*3)+(1090*2)+(1110*4)+(1130*0)+(1150*2)+(1170*1)+(1190*1)+(1210*0)+(1230*1)= 40700

X[bar]= \frac{40700}{40} = 1017.5

Mo= 1012 < Me= 1013.33 < X[bar]= 1017.5

Looking only at the measurements of central tendency you could wrongly conclude that the distribution is symmetrical or slightly skewed to the right since the three values are included in the same interval but not the same number.

*-*-*

Now you have to do the same but changing the class with (interval amplitude) to 100, starting at 800

Example

1) [800;900) There are 4 observations that are included in this interval: 819, 836, 888, 897 , so f1=4

2)[900;1000) There are 12 observations that are included in this interval: 903, 907, 912, 918, 942, 943, 952, 959, 962, 986, 992, 994 , so f2= 12

etc...

As you can see this distribution is more uniform, increasing the amplitude of the intervals not only decreased the number of class intervals but now we observe that there are observed frequencies for all of them.

Mode:

The largest absolute frequency is f(3)=15, so the mode interval is [1000;1100)

Using the same formula as before:

Mo= 1000 + 100*(\frac{(15-12)}{(15-12)+(15-8)} )=1030

This distribution is unimodal.

Median:

Position of the median n/2= 40/2= 20

As before is the 20th observed frequency, this frequency is included in the interval [1000;1100)

Me= 1000+ 100*[\frac{20-16}{15} ]= 1026.67

Mean:

∑x'*fi= (850*4)+(950*12)+(1050*15)+(1150*8)+(1250*1)= 41000

X[bar]= \frac{41000}{40} = 1025

X[bar]= 1025 < Me= 1026.67 < Mo= 1030

The three values are included in the same interval, but seeing how the mean is less than the median and the mode, I would say this distribution is symmetrical or slightly skewed to the left.

I hope it helps!

8 0
4 years ago
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