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Tom [10]
3 years ago
10

A triangle is 3cm wider than it is tall. The area is 44cm^2. Find the width(length of the base).

Mathematics
1 answer:
stich3 [128]3 years ago
8 0
Set height to x, then length of base is x+3, Area formula says that
0.5*x*(x+3)=44, x(x+3)=88, x=8, x+3=11, which is the length you are looking for (11cm)
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a straight line has a equation 4x + 3y = 12 . find the gradient of the line and the coordinates of the point where this line cro
Komok [63]
4x + 3y = 12
3y = -4x + 12
y = -4/3x + 4........so the slope (or gradient) is -4/3...because in y = mx + b form, the slope(gradient) is in the m position and the y int is in the b position....so if u wanted to know the y axis, it would be (0,4)

the x intercept (where the line crosses the x axis) can be found by subbing in 0 for y in the original equation or the slope intercept equation, and solving for x.
4x + 3(0) = 12
4x = 12
x = 12/4 = 3....so the x intercept is (3,0)
4 0
3 years ago
Factor 25 into prime numbers
kirill [66]

Image result for factor 25 into prime numbers

25 is a composite number, and it is 5 squared. 25 = 1 x 25 or 5 x 5. Factors of 25: 1, 5, 25. Prime factorization: 25 = 5 x 5, which can also be written 25 = 5².

8 0
3 years ago
What is 3/8 x 16/7? i don't know. it is for my math homework?
liberstina [14]
Multiply the top and bottom.

3/8 * 16/7 = 48/56

Simplified: 5/7
6 0
3 years ago
Determine the equations of the vertical and horizontal asymptotes, if any, for y=x^3/(x-2)^4
djverab [1.8K]

Answer:

Option a)

Step-by-step explanation:

To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

\lim_{x\to\\2}\frac{x^3}{(x-2)^4} \\\\\\lim_{x\to\\2}\frac{2^3}{(2-2)^4}\\\\\lim_{x\to\\2}\frac{2^3}{(0)^4} = \infty

Then. x = 2 it's a vertical asintota.

To obtain the horizontal asymptote of the function take the following limit:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4}

if \lim_{x \to \infty}\frac{x^3}{(x-2)^4} = b then y = b is horizontal asymptote

Then:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4} \\\\\\lim_{x \to \infty}\frac{1}{(\infty)} = 0

Therefore y = 0 is a horizontal asymptote of f(x).

Then the correct answer is the option a) x = 2, y = 0

3 0
3 years ago
Read 2 more answers
heather has divided $6300 between two invesments, one paying 9%, the other paying 4%. If the return on her investment is $372, h
Ganezh [65]
Assuming that the two investments are  X & Y
X + Y = 6300
X = 6300 - Y                                       (1)

9/100X + 4/100 Y = 372                       (2)
replacing X from (1) into (2)
9/100(6300-Y) + 4/100 Y = 372
567 - 9/100Y +4/100Y = 372
(-9+4)/100Y = 372 - 567
5/100Y = -195
Y = 100*195/5 = 3900
From (1) we can get X
X = 6300 - 3900 = 2400

I hope this is helpful 



6 0
3 years ago
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