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Morgarella [4.7K]
3 years ago
13

What is the area of a triangle with base of 20 feet and a vertical height

Mathematics
1 answer:
VashaNatasha [74]3 years ago
8 0

Answer:

A = 400 ft²

Step-by-step explanation:

The area (A) of a triangle is calculated as

A = \frac{1}{2} bh ( b is the base and h the vertical height ) , then

A = \frac{1}{2} × 20 × 40 = 10 × 40 = 400 ft²

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Make 'y' the subject of the equation. g = 3y - 9.
Vikentia [17]

Answer:

y = \frac{g+9}{3}

Step-by-step explanation:

g = 3y -9

start by adding 9 to each side:

g + 9 = 3y

now divide by 3 on each side to get y alone:

y = \frac{g+9}{3}

hope this helps!

6 0
3 years ago
Write an equation for each line. y intercept of -2,1, x intercept of 3.5
grandymaker [24]
Y = -2.1 , x = 3.5 the first line will be horizontal and the other will be vertical
8 0
3 years ago
7n+12=1/2(14n+24)<br><br><br> NEED ANSWER ASAP
Angelina_Jolie [31]

Answer:

n = all real numbers

Step-by-step explanation:

7n+12=1/2(14n+24)

multiply each side by 2

2(7n+12)=2*1/2(14n+24)

distribute

14n+24 = 14n + 24

subtract 14n from each side

14n-14n+24 = 14n-14n + 24

24=24

this is always true

n = all real numbers

3 0
3 years ago
Which is the value of i^20+1?
Nonamiya [84]

Answer:

Step-by-step explanation:

i^{20}+1=i^{2*10}+1\\\\(i^{2})^{10}+1=(-1)^{10}+1\\\\=1+1=2

8 0
3 years ago
Which graph represents the function f (x) = StartFraction 2 Over x minus 1 EndFraction + 4?
Pepsi [2]

Answer:

On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4

Step-by-step explanation:

The given function is presented as follows;

f(x) = \dfrac{2}{x - 1} + 4

From the given function, we have;

When x = 1, the denominator of the fraction, \dfrac{2}{x - 1}, which is (x - 1) = 0, and the function becomes, \dfrac{2}{1 - 1} + 4 = \dfrac{2}{0} + 4 = \infty + 4 = \infty therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote

Also we have that in the given function, as <em>x</em> increases, the fraction \dfrac{2}{x - 1} tends to 0, therefore as x increases, we have;

\lim_  {x \to \infty}  \dfrac{2}{(x - 1)} \to 0, and \  \dfrac{2}{(x - 1)}  + 4 \to 4

Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1

When -∞ < x < 1, we also have that as <em>x</em> becomes more negative, f(x) → 4. When x = 0, \dfrac{2}{0 - 1} + 4 = 2. When <em>x</em> approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left

Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.

5 0
2 years ago
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