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myrzilka [38]
3 years ago
9

Find the area of this triangle.

Mathematics
1 answer:
Misha Larkins [42]3 years ago
6 0

Answer:

72 m

Step-by-step explanation:

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one positive number is 12 times another positive number. The difference between the two numbers is 143
Natali [406]
Let x be the unknown number and let's set up an equation.
12x - x = 143
Now solve for x:
11x = 143
x = 13
The positive number is 13.
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250% of 9.4 mi is what distance
slava [35]

Answer: 23.5m

Step-by-step explanation:

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3 years ago
This question has three parts. Answer the parts in order.
nalin [4]

Answer:

The area of the smallest section is A_{1}=100yd^{2}

The area of the largest section is A_{2}=625yd^{2}

The area of the remaining section is A_{3}=250yd^{2}

Step-by-step explanation:

Please see the picture below.

1. First we are going to name the side of the larger square as x.

As the third section shares a side with the larger square and the four sides of a square are equal, we have the following:

- Area of the first section:

A_{1}=10yd*10yd

A_{1}=100yd^{2}

- Area of the second section:

A_{2}=x^{2} (Eq.1)

- Area of the third section:

A_{3}=width*length

A_{3}=10yd*x (Eq.2)

2. The problem says that the total area of the enclosed field is 975 square yards, and looking at the picture below, we have:

A_{1}+A_{2}+A_{3}=975yd^{2}

Replacing values:

100+x^{2}+10x=975

Solving for x:

x^{2}+10x-875=0

x=\frac{-10+\sqrt{100+(4*875)}}{2}

x=\frac{-10+\sqrt{3600}}{2}

x=\frac{-10+60}{2}

x=25

3. Replacing the value of x in Eq.1 and Eq.2:

- From Eq.1:

A_{2}=25^{2}

A_{2}=625yd^{2}

- From Eq.2:

A_{3}=10*25

A_{3}=250yd^{2}

3 0
2 years ago
Solve the following problems using 5-D process part 2
svet-max [94.6K]

The two problems are almost identical: you have to set up a system, and then solve it.

In the first case, let g and b be, respectively, the number of girls and boys.

We know that g=2b+4 (the number of girls in the Spanish Club is four more than twice the number of boys). Also, we know that g+b=61 (there are 61 students in total).

So, we have the system

\begin{cases}g=2b+4\\g+b=61\end{cases}

We can use the first equation to substitute in the second

g+b=61 \iff (2b+4)+b=61 \iff 3b+4=61 \iff 3b=57 \iff b=19

And then solve for g:

g=2b+4=2\cdot 19+4=38+4=42

For the second problem, let c and j be the number of pins knocked down by, respectively, Carrie and John. Just like before, we have the system

\begin{cases}c=j+12\\c+j=230\end{cases}

And you can solve it in the very same way we solved the previous one.

5 0
3 years ago
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