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3 years ago

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The game room of the youth center needs carpet. sweet t found that the room measures 12 inches by 8 inches on the blueprint. if

the scale of the blueprint is 1 inch to 3 feet, what is the actual area of the game room.

1 answer:

mojhsa [17]3 years ago

8 0

The area of the game room is 864 square feet

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A ball is thrown into the air. The height in feet of the ball can be modeled by the equation h = -16t2 + 20t + 6 where t is the

makkiz [27]

$f(t)=h=-16t^2+20t+6

$

a=-16, b=20, c =6

Δ= $b^2-4ac=20^2-4*(-16)*6=400+384=784$

$\sqrt{Δ}=\sqrt{784}=28$

max:

$p=-\frac{b}{2a}=-\frac{20}{-32}=\frac{20}{32}=\frac{5}{8}$

max heigh: $h=-16(\frac{5}{8})^2+20*\frac{5}{8}+6=-16\frac{25}{64}+\frac{100}{8}+6$

$h=-\frac{25}{4}+\frac{50}{4}+6=\frac{-25+50+24}{4}=\frac{49}{4}$

$h=\frac{49}{4}$ at $t=\frac{5}{8}s$

will fall down at t:

$t=\frac{-20-28}{-32}=\frac{-48}{-32}=\frac{3}{2}=1.5$

$t=1.5s$

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3 years ago

Question 4 of 10

ludmilkaskok [199]

Answer:

Option C. 3√3

Step-by-step explanation:

Please see attached photo for brief explanation.

In the attached photo, we obtained the following:

Opposite = a

Adjacent = 3

Hypothenus = 6

Angle θ = 60°

We can obtain the value of 'a' as follow:

Tan θ = Opposite /Adjacent

Tan 60° = a/3

Cross multiply

a = 3 x Tan 60°

But: Tan 60° = √3

a = 3 x Tan 60°

a = 3 x √3

a = 3√3

Therefore, the length of the altitude of the equilateral triangle is 3√3.

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3 years ago

There are 20 girls on the basketball team. of these, 17 are over 16 years old, 12 are taller than 170 cm, and 9 are both older t

Oduvanchick [21]

17 are older than 16 years old and 12 are taller than 170 cm

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3 years ago

A wedding planner purchased both small and large lanterns for a wedding reception. The planner purchased a total of 40 lanterns

Olegator [25]

Answer:

The wedding planner must have purchased 28 small lanterns and 12 large lanterns.

Step-by-step explanation:

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2 years ago

Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S

Sophie [7]

Answer:

Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that $A=J^{-1}AJ$ . Thus, A↔A.

<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that $A=P^{-1}BP$ . In this equality we can perform a right multiplication by $P^{-1}$ and obtain $AP^{-1} =P^{-1}B$ . Then, in the obtained equality we perform a left multiplication by P and get $PAP^{-1} =B$ . If we write $Q=P^{-1}$ and $Q^{-1} = P$ we have $B = Q^{-1}AQ$ . Thus, B↔A.

<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have $A=P^{-1}BP$ and from B↔C we have $B=Q^{-1}CQ$ . Now, if we substitute the last equality into the first one we get

$A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP)$ .

Recall that if P and Q are invertible, then QP is invertible and $(QP)^{-1}=P^{-1}Q^{-1}$ . So, if we denote R=QP we obtained that

$A=R^{-1}CR$ . Hence, A↔C.

Therefore, the relation is an <em>equivalence relation</em>.

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3 years ago