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gtnhenbr [62]
2 years ago
8

The diagram shows several points and lines.

Mathematics
1 answer:
Ghella [55]2 years ago
4 0

Answer:

Option B And Option E

Step-by-step explanation:

three or more points  are said to be collinear if they lie on a single straight line.

In the given figure :

Points J, K, and Q are collinear.

Through any two given  points only one line can be drawn.

In the given figure:

There is only one line that can be drawn through points L and P.

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An arena receives $1,700 per event for 12 concerts. If the costs for this sponsorship total $14,000, what is the profit margin f
Jobisdone [24]

Answer:

The profit margin for the concert sponsorship is 31.3725 %

Step-by-step explanation:

An arena receives for 1 event = $1700

An arena receives for 12 events = 12 \times 1700

An arena receives for 12 events = 20400

The costs for this sponsorship total $14,000

Profit = 20400-14000

Profit =6400

Profit margin =\frac{6400}{20400} \times 100

Profit margin = 31.3725\%

Hence the profit margin for the concert sponsorship is 31.3725 %

6 0
1 year ago
The histogram to the right represents the weights​ (in pounds) of members of a certain​ high-school programming team. What is th
AVprozaik [17]

Answer:

Class width=20

Lower class limit of first class=100

Upper class limit of first class=120

Step-by-step explanation:

The class width can be calculated by taking the difference of two consecutive upper class limits or lower class limits.

Now we take any two consecutive upper class limits or lower class limits from classes 100-120,120-140,140-160,160-180,180-200,200-220,220-240.

We take upper class limits of first and second class i.e. 100 and 120.

Class width=120-100=20

Class width=20

The approximate lower and upper class limits of the first class from classes  100-120,120-140,140-160,160-180,180-200,200-220,220-240 are 100 and 120.

Class limits for first class is 100-120.

Lower class limit of first class=100

Upper class limit of first class=120

4 0
1 year ago
30 POINTS A quadrilateral has vertices A(11, -7), B(9, -4), C(11, -1), and D(13, -4). Quadrilateral ABCD is a __________ A.paral
Otrada [13]

Answer:

Rombus  which is D and then B

Step-by-step explanation:

If you go to desmos (website) they have a graphing calculator which allows you to graph how the shapes will look like. The first quadrilateral makes a rombus as it is a sideways square and the second quadrilateral is a kite as the rombus top point is stretched upwards.

9 0
2 years ago
. Solve the following initial value problem: (t2−20t+51)dydt=y (t2−20t+51)dydt=y with y(10)=1y(10)=1. (Find yy as a function of
Semenov [28]

Answer:

y=(\frac{t-17}{t-3})^{\frac{1}{14}}

Step-by-step explanation:

We are given that initial value problem

t^2-20t+51)\frac{dy}{dt}=y

\frac{dy}{y}=\frac{dt}{t^2-20t+51}

\frac{dy}{y}=\frac{dt}{t^2-3t-17t+51}

\frac{dy}{y}=\frac{dt}{(t(t-3)-17(t-3)}

\frac{dy}{y}=\frac{dt}{(t-3)(t-17)}

\frac{1}{(t-3)(t-17)}=\frac{A}{t-3}+\frac{B}{t-17}

\frac{1}{(t-3)(t-17)}=\frac{A(t-17)+B(t-3)}{(t-3)(t-17)}

1=A(t-17)+B(t-3)...(1)

Substitute t-3=0

t=3

t-17=0

t=17

Substitute t=3 in equation (1)

1=A(3-17)+0

1=-14A

A=-\frac{1}{14}

Substitute t=17

1=B(17-3)

1=14B

B=\frac{1}{14}

Substitute the values of A and B

\frac{1}{(t-3)(t-17)}=-\frac{1}{14}(\frac{1}{t-3})+\frac{1}{14}(\frac{1}{t-17})

\int\frac{dy}{y}=-\frac{1}{14}\int\frac{dt}{t-3}+\frac{1}{14}\int\frac{dt}{t-17}

ln y=-\frac{1}{14}ln\mid{t-3}\mid+\frac{1}{14}\mid{t-17}\mid+ln C

By using formula:\frac{dx}{x}=ln x+C

ln y=\frac{1}{14}(-ln\mid{t-3}\mid+ln\mid{t-17}\mid)+ln C

Using formula:ln x-ln y=ln \frac{x}{y}

ln y=\frac{1}{14}(ln\mid{\frac{t-17}{t-3}}\mid)+ln C

ln y=\frac{1}{14}ln\mid{\frac{t-17}{t-3}}\mid+ ln C

Substitute y(10)=1

ln 1=\frac{1}{14}ln\mid\frac{10-17}{10-3}\mid+ln C

0=0+ln C

Because ln 1=0

lnC=0

C=e^0=1

Because ln x=y\implies x=e^y

Substitute the value of C

ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid+ln1

ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid+0

ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid

14ln y=ln\mid\frac{t-17}{t-3}\mid

lny^{14}=ln\mid\frac{t-17}{t-3}\mid

By using identity blog a= loga^b

y^{14}=\frac{t-17}{t-3}

y=(\frac{t-17}{t-3})^{\frac{1}{14}}

6 0
2 years ago
Prove the diagonals of a parallelogram bisect one another. Be sure to create and name the appropriate geometric figures.
GalinKa [24]
Hope this helps you!!!...

3 0
2 years ago
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