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Irina-Kira [14]
3 years ago
12

The altitude of a triangle is 5 less than its base. The area of the triangle is 42 square inches. Find its base and altitude.​

Mathematics
1 answer:
zavuch27 [327]3 years ago
8 0

9514 1404 393

Answer:

  base: 12 in

  altitude: 7 in

Step-by-step explanation:

The formula for the area of a triangle is ...

  A = 1/2bh

Filling in the given information, we have ...

  42 = 1/2(b)(b -5)

  b^2 -5b -84 = 0 . . . . multiply by 2, put in standard form

  (b -12)(b +7) = 0 . . . . factor

The only positive solution is b = 12

The base is 12 inches; the altitude is 7 inches.

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notsponge [240]

Answer:

C.

Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
The length of a rectangle is 8 feet more than its width. If the width is increased by 4 feet and the length is decreased by 5 fe
bezimeni [28]
Assign the following variables for the origina3l rectangle:
let w = width let w + 8 = length and the area would be w(w + 8) = w² + 8w

No for the second rectangle:
let (w + 4) = width and (w + 8 - 5) or (w + 3) = length
Area = length x width or (w + 4)(w + 3) = w² + 3w + 4w + 12 using the foil method to multiply to binomials. Simplified Area = w² + 7w + 12

Now our problem says that the two area will be equal to each other, which sets up the following equation:

w² + 8w = w² + 7w + 12 subtract w² from both sides
8w = 7w + 12 subtract 7w from both sides
w = 12 this is the width of our original rectangle
recall w + 8 = length, so length of the original rectangle would be 20
7 0
3 years ago
Solve for the variable a D =a+b-c/3
yuradex [85]
Hey there,
solving for a is just like you think of all other letter as a constant
i.e the only variable is a
so you need to subtract b as a start
      d=a+b-c/3
-b             -b
=============

    -b+d=a-c/3                now it's time to add c/3 (isolate a by itself in one side)

     -b+d=a-c/3
+c/3         +c/3
==============
c/3-b+d=a

Hence you have solved for a




4 0
4 years ago
I know that real numbers consist of the natural or counting numbers, whole numbers, integers, rational numbers and irrational nu
ra1l [238]

The imaginary unit i belongs to the set of complex numbers, denoted by \mathbb C. These numbers take the form a+bi, where a,b are any real numbers.

The set of real numbers, \mathbb R, is a subset of \mathbb C, where each number in \mathbb R can be obtained by taking b=0 and letting a be any real number.

But any number in \mathbb C with non-zero imaginary part is not a real number. This includes i.

  • "is it possible that i can use an imaginary number for a real number"

I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because 2=2+0i.

There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising i to the i-th power. Since i=e^{i\pi/2}, we have

i^i=\left(e^{i\pi/2}\right)^i=e^{i^2\pi/2}=e^{-\pi/2}\approx0.2079

3 0
3 years ago
A water tower casts a 100-foot shadow. At the same time, an 8-foot stri
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Set up ratios of height / shadow length:

8/6 = x/100

Cross multiply

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Divide both sides by 6

X = 133.33 feet

The tower is 133.3 feet

6 0
3 years ago
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