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BigorU [14]
3 years ago
10

First to answer right gets brainliest!!! You are making a mosaic design on a square table top. You have already covered half of

the table top with 150 1-inch square tile pieces.
A. What are the dimensions of the table top?

B. What is the measure of the diagonal from one corner to the opposite corner of the table top?
Mathematics
1 answer:
lapo4ka [179]3 years ago
8 0

Answer:

i don't know

Step-by-step explanation:

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What is the area of the figure:
Leni [432]
A <span>45 45 90 triangle
so a = b
 a^2 + b^2 = </span>c^2<span>
 a^2 + a^2 </span>= c^2<span>
 2a^2 </span>= c^2<span>
 2a^2 </span>= 24^2
<span> 2a^2 = 576
   a^2 = 288
       a = 12</span>√2
 b = a = 12√2
<span>
Area of triangle:
= 1/2 (</span>12√2)(12√2)
<span>= 1/2(288)
= 144

answer
Area = 144 ft^2

</span>
8 0
3 years ago
Round to the nearest hundredth 0.633
spayn [35]
.63 i mean it is not that hard

6 0
3 years ago
Read 2 more answers
Which of these describes two thirds?​
vladimir1956 [14]

Answer:

2/3

two thirds

0.666666666

Step-by-step explanation:

please mark this answer as brainliest

5 0
3 years ago
A Rectangle with sides 100m and 50m. Person B and Person A start walking in the opposite direction at the same point with veloci
Maurinko [17]

Answer:

It takes takes them 31.67 secs to meet after Person A starts walking

OR

34.67 secs after Person B starts walking

Step-by-step explanation:

First, we will find the perimeter of the rectangle

From,

Perimeter of rectangle = 2 (l + b)

Where l is the length

and b is the breadth

From the question, l = 100m

and b = 50m

Hence,

Perimeter of the rectangle = 2 (100+50)

= 2(150) = 300m

Hence, the perimeter of the rectangle is 300m

This is the total distance that will be covered by both persons by the time they meet.

Let the distance covered by person A be S_{1}

and the distance covered by person B be S_{2}

We can write that

S_{1} + S_{2} = 300m

Velocity of person A, V_{A} = 4 m/s

and velocity of person B, V_{B} = 5 m/s

From the question, Person A starts walking 3 seconds after person B,

This means, if person A spends t secs before they meet, then person B would spend (3 + t) secs.

For Person A,

Velocity = 4 m/s

Time = t secs

Distance = S_{1}

From,

Velocity = \frac{Distance}{Time}

Then,

Distance = Velocity \times time

S_{1} = 4 \times t

S_{1} = 4t ...... (1)

For Person B

Velocity = 5 m/s

Time = (t + 3) secs

Distance = S_{2}

Also, from

Distance = Velocity \times time

S_{2} = 5 \times (3+t)

S_{2} = 5(3+t) ...... (2)

Recall that, S_{1} + S_{2} = 300m

Then, S_{2} = 300m - S_{1}

We can then write that,

300m - S_{1} = 5(3+t)

Then,

S_{1} = 300 - 5(3+t) ..... (3)

Equating equations (1) and (3), we get

300 - 5(3+t) = 4t

300 - 15 -5t = 4t\\9t = 285\\t = \frac{285}{9}\\

t = 31.67 secs

This is the time spent by Person A

Hence, it takes takes them 31.67 secs to meet after Person A starts walking OR

34.67 secs after Person B starts walking

3 0
3 years ago
PLEASE HELP ANSWER THESE QUESTIONS!
Sliva [168]

Answer:

Step-by-step explanation:

a² - b² = (a+ b)(a - b)

1) (2n-4/2n) ÷ (n^2-4/n)

=\frac{2n-4}{2n}*\frac{n}{n^{2}-4}\\\\=\frac{2n-2*2}{2n}*\frac{n}{n^{2}-2^{2}}\\\\=\frac{2*(n-2)}{2n}*\frac{n}{(n+2)*(n-2)}\\\\=\frac{1}{n+2}

2) [y^2-36/y^2-49] ÷[ y+6/y-7]

=\frac{y^{2}-36}{y^{2}-49}*\frac{y-7}{y+6}\\\\=\frac{y^{2}-6^{2}}{y^{2}-7^{2}}*\frac{y-7}{y+6}\\\\=\frac{(y+6)*(y-6)}{(y+7)*(y-7)}*\frac{y-7}{y+6}\\\\=\frac{y-6}{y+7}\\

3) [m^2-1/ m^2-m] ÷ [m^2-7m-8/3m ]

=\frac{m^{2}-1}{m^{2}-m}*\frac{3m}{m^{2}-7m-8}\\\\=\frac{(m+1)*(m-1)}{m*(m-1)}*\frac{3m}{(m-8)*(m+1)}\\\\=\frac{3}{m-8}

Hint :  m² - 7m - 8

sum = -7

Product  = -8

Factor = (-8), 1

m² - 7m - 8  =m² - 8m + m - 8

                   = m*(m - 8) + (m-8)

                   = (m - 8)(m +1)

7 0
3 years ago
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