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3 years ago

12

At Cooper High School, jazz band is only offered as a zero-hour (before school) class. Dan has a zero-hour class. Conjecture: Da

n is in jazz band. Use deductive reasoning to verify the conjecture, or provide a counterexample if the conjecture is false.

1 answer:

Helga [31]3 years ago

5 0

false; Dan may be enrolled in a different zero-hour class

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One week, Carlos bought 3 bags of Tabitha Tidbits and 4 bags of Figaro Flakes for $43.00. The next week he bought 3 bags of Tabi

andrezito [222]

The cost of one bag of Tabitha Tidbits costs $7.00 and cost of one bag of Figaro Flakes costs $5.50

Step-by-step explanation:

Let,

Cost of one bag of Tabitha Tidbits = x

Cost of one bag of Figaro Flakes = y

According to given statement;

3x+4y=43.00 Eqn 1

3x+6y=54.00 Eqn 2

Subtracting Eqn 1 from Eqn 2

$(3x+6y)-(3x+4y)=54.00-43.00\\3x+6y-3x-4y=11.00\\2y=11.00$

Dividing both sides by 2

$\frac{2y}{2}=\frac{11}{2}\\y=5.50$

Putting y=5.50 in Eqn 1;

$3x+4(5.50)=43.00\\3x+22.00=43.00\\3x=43.00-22.00\\3x=21.00$

Dividing both sides by 3

$\frac{3x}{3}=\frac{21.00}{3}\\x=7.00$

The cost of one bag of Tabitha Tidbits costs $7.00 and cost of one bag of Figaro Flakes costs $5.50

Keywords: linear equations, elimination method

Learn more about elimination method at:

#LearnwithBrainly

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3 years ago

Finding a Product

Vinil7 [7]

Answer: A & C

Step-by-step explanation:

Got it on edgenuity

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4 years ago

What is the area and perimeter of this shape? I need help!!

cluponka [151]

The area of that shape is: 200 cm

The perimeter of that shape is: 72 cm

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3 years ago

Read 2 more answers

Stella [2.4K]

Check the picture below, so the park looks more or less like so, with the paths in red, so let's find those midpoints.

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad K(\stackrel{x_2}{1}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 1 -3}{2}~~~ ,~~~ \cfrac{ 3 +1}{2} \right) \implies \left(\cfrac{ -2 }{2}~~~ ,~~~ \cfrac{ 4 }{2} \right)\implies JK=(-1~~,~~2) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ L(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 +5}{2}~~~ ,~~~ \cfrac{ -3 -1}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ -4 }{2} \right)\implies LM=(2~~,~~-2) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JK(\stackrel{x_1}{-1}~,~\stackrel{y_1}{2})\qquad LM(\stackrel{x_2}{2}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JKLM=\sqrt{(~~2 - (-1)~~)^2 + (~~-2 - 2~~)^2} \\\\\\ JKLM=\sqrt{(2 +1)^2 + (-2 - 2)^2} \implies JKLM=\sqrt{( 3 )^2 + ( -4 )^2} \\\\\\ JKLM=\sqrt{ 9 + 16 } \implies JKLM=\sqrt{ 25 }\implies \boxed{JKLM=5}$

now, let's check the other path, JM and KL

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 -3}{2}~~~ ,~~~ \cfrac{ -3 +1}{2} \right) \implies \left(\cfrac{ -4 }{2}~~~ ,~~~ \cfrac{ -2 }{2} \right)\implies JM=(-2~~,~~-1) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ K(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad L(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 5 +1}{2}~~~ ,~~~ \cfrac{ -1 +3}{2} \right) \implies \left(\cfrac{ 6 }{2}~~~ ,~~~ \cfrac{ 2 }{2} \right)\implies KL=(3~~,~~1) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JM(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad KL(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JMKL=\sqrt{(~~3 - (-2)~~)^2 + (~~1 - (-1)~~)^2} \\\\\\ JMKL=\sqrt{(3 +2)^2 + (1 +1)^2} \implies JMKL=\sqrt{( 5 )^2 + ( 2 )^2} \\\\\\ JMKL=\sqrt{ 25 + 4 } \implies \boxed{JMKL=\sqrt{ 29 }}$

so the red path will be $5~~ + ~~\sqrt{29} ~~ \approx ~~ \blacksquare~~ 10 ~~\blacksquare$

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2 years ago

A baseball player recorded 40 hits in 166 at-bats. What is the ratio of hits to at-bats?

Likurg_2 [28]

Ratio = 40 : 166 = 20 : 83

Answer: 20 : 83

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4 years ago