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raketka [301]
3 years ago
5

ABC and ADC are triangles. The area of triangle ADC is 52m^2

Mathematics
1 answer:
sweet [91]3 years ago
4 0

Given that,

ABC and ADC are triangles.

The area of ΔADC is 52 m².

Suppose , AD is the median.

According to figure,

We need to find the area of ΔABC

Using theorem of triangle

\bigtriangleup ADB +\bigtriangleup ADC=\bigtriangleup ABC

Here, Δ ADB = Δ ADC

So, 2 \bigtriangleup ADC=\bigtriangleup ABC

Put the value of Δ ADC

\bigtriangleup ABC =2\times52

\bigtriangleup ABC = 104\ m^2

Hence, The area of ΔABC is 104 m².

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Crazy boy [7]

Answer:

1/3

Step-by-step explanation:

To get no solution, you want to get rid of x and leave an incorrect equation.

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1/3x + 2 = 1/3x - 3

We subtract 1/3x on both sides. This will cause the 1/3x's to cancel out.

1/3x - 1/3x + 2 = 1/3x - 1/3x - 3

2 ≠ -3

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Kate has 21 coins (nickels and dimes) in her purse. How many nickels and dimes does she have if she has $1.50? 3 nickels and 18
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4) Adult tickets for the evening show at XD Theater are $11 and children's tickets are $Belf the theater seats 250 people, how m
kozerog [31]

Omitted value: The price of children ticket was omitted in the question, so i used $8 to solve. You can input the correct value and solve the same way following the steps.

Answer: 100 adult tickets  must be sold.

Step-by-step explanation:

step 1

let  x represent Adults

AND y represent children

Since the theater seats 250 people we have that

x+y = 250..... equation 1

Also price for Adult ticket = $11

and  price children ticket =$8

With total sales at $2,300, we have that

11x + 8y= 2300----- equation 2

Step 2

Making y subject in equation 1

' x+y = 250

y= 250-x

Putting y= 250- x in equation 2

11x + 8(250-x)= 2300

11x +2000-8x= 2300

11x -8x = 2300-2000

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x 300/3

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 x+y = 250

100+y=250

y=250-100

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5 0
3 years ago
❊ Simplify :
DiKsa [7]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to simplify:

\displaystyle \frac{a+2}{a^2+a-2}+\frac{3}{a^2-1}

First, let's factor the denominators of each term. For the second term, we can use the difference of two squares. Hence:

\displaystyle =\frac{a+2}{(a+2)(a-1)}+\frac{3}{(a+1)(a-1)}

Now, create a common denominator. To do this, we can multiply the first term by (<em>a</em> + 1) and the second term by (<em>a</em> + 2). Hence:

\displaystyle =\frac{(a+2)(a+1)}{(a+2)(a-1)(a+1)}+\frac{3(a+2)}{(a+2)(a-1)(a+1)}

Add the fractions:

\displaystyle =\frac{(a+2)(a+1)+3(a+2)}{(a+2)(a-1)(a+1)}

Factor:

\displaystyle =\frac{(a+2)((a+1)+3)}{(a+2)(a-1)(a+1)}

Simplify:

\displaystyle =\frac{a+4}{(a-1)(a+1)}

We can expand. Therefore:

\displaystyle =\frac{a+4}{a^2-1}

Problem 2)

We want to simplify:

\displaystyle \frac{1}{(a-b)(b-c)}+\frac{1}{(c-b)(a-c)}

Again, let's create a common denominator. First, let's factor out a negative from the second term:

\displaystyle \begin{aligned} \displaystyle &= \frac{1}{(a-b)(b-c)}+\frac{1}{(-(b-c))(a-c)}\\\\&=\displaystyle \frac{1}{(a-b)(b-c)}-\frac{1}{(b-c)(a-c)}\\\end{aligned}

Now to create a common denominator, we can multiply the first term by (<em>a</em> - <em>c</em>) and the second term by (<em>a</em> - <em>b</em>). Hence:

\displaystyle =\frac{(a-c)}{(a-b)(b-c)(a-c)}-\frac{(a-b)}{(a-b)(b-c)(a-c)}

Subtract the fractions:

\displaystyle =\frac{(a-c)-(a-b)}{(a-b)(b-c)(a-c)}

Distribute and simplify:

\displaystyle =\frac{a-c-a+b}{(a-b)(b-c)(a-c)}=\frac{b-c}{(a-b)(b-c)(a-c)}

Cancel. Hence:

\displaystyle =\frac{1}{(a-b)(a-c)}

4 0
3 years ago
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