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Ahat [919]
3 years ago
8

Find the perimeter and area of the figure, Explain your answer.

Mathematics
1 answer:
mel-nik [20]3 years ago
4 0
5.5 + 5.7 + 9 yards = 20.2 yards for perimeter
The area is 15 yards.
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That’s what I need to know to set it up plzzzzss help
AURORKA [14]
3×(60+7)
(3×30) + (3×7)
90 + 21 = 111
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Solve the equation.<br> -1/6x - 5 = 2/3x<br> Please type step-by-step explanation.
Debora [2.8K]

Answer: Solution

=

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6

Step-by-step explanation:

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I need help with this Maths Homework Question.
Monica [59]

Answer:

1. 41/45.

2. x/(x^2+3x+6)

Step-by-step explanation:

1.

So first we fill the ven diagram.

There are 240 in band, so we fill that. 60 students are in both, we put that in the middle, and there are 110 people in choir.

now, since we want the probability that a student is chosen that is in band, and choir, and both. We add all this up

240 + 60 + 110 = 410.

The total possible outcome is 410, and the total outcome is 450, so the answer is

410/450 = 41/45

2.

First, to get the total outcomes, we have to add all the expressions together.

x(x-2) + x + 2x+8 = x^2 - 2 + x + 2x + 8 = x^2 - 2 + 3x + 8 = x^2 + 3x + 6.

Since that is the total outcome, we have to find the possible outcomes.

The problem wants BOTH from the 20th century and British, so it is x.

x/(x^2+3x+6). We cannot simplify any further, thus x/(x^2+3x+6) is our answer

3 0
2 years ago
How much square units of 50 units
Alborosie

Answer:

7.07106781187

Step-by-step explanation:

if its just asking \sqrt50

:D

5 0
3 years ago
Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and
Lisa [10]

Answer:

Given :

ABC is a right triangle in which ∠ABC = 90°,

Also, Legs AB and CB are extended past point B to points D and E,

Such that,

\angle EAC = \angle ACD = 90^{\circ}

To prove :

EB\times BD=AB\times BC

Proof :

In triangles AEC and EBA,

∠EAC= ∠ABE ( right angles )

∠CEA = ∠AEB ( common angles )

By AA similarity postulate,

\triangle AEC \sim \triangle EBA,

Similarly,

\triangle AEC \sim \triangle ABC

\implies\triangle EBA\sim \triangle ABC-----(1)

Now, In triangles ADC and CBD,

∠ACD = ∠CBD ( right angles )

∠ADC= ∠BDC ( common angles )

By AA similarity postulate,

\triangle ADC \sim \triangle CBD,

Similarly,

\triangle ADC \sim \triangle ABC

\implies \triangle CBD\sim \triangle ABC-----(2)

From equations (1) and (2),

\triangle EBA\sim \triangle CBD

The corresponding sides of similar triangles are in same proportion,

\frac{EB}{BC}=\frac{AB}{BD}

\implies EB\times BD=AB\times BC

Hence, proved....

5 0
3 years ago
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