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Andrei [34K]
3 years ago
14

Find the limit of the function by using direct substitution. limit as x approaches four of quantity x squared plus three x minus

one
Mathematics
2 answers:
MAXImum [283]3 years ago
8 0

Answer:

The correct answer is 27, when using direct substitution

Step-by-step explanation:

(4)^2+3(4)-1 --> 16+12-1 --> 16+11= 27

Yuliya22 [10]3 years ago
5 0
\lim_{x \to 4} f(x)=x^2+3x-1
just subsitute
f(4)=4²+3(4)-1
f(4)=16+12-1
f(4)=28-1
f(4)=27

it approaches 27 as x approaches 4
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The vertices of a hyperbola are located at (−4, 1) and (4, 1). The foci of the same hyperbola are located at (−5, 1) and (5, 1).
topjm [15]

Answer:

x^2\ 16 - y^2/9 = 1

step-by-step explanation:

soooo the equation for a hyperbola that is horizontal is x^2/a^2 - y^2/b^2 = 1

a hyperbola should always equal one so dont forget that when writing your equation becuase it is easy to forget.

it helps to graph this so you can see it better

to find a it is the distance from the center to the vertices which is 4 so in the equation you will write 16 becuase it is a^2

then you need to find b. to get b you have to figure out that c is the distance from the center to the foci which is 5 and it is all related to the plythagorm theorum becuase it forms a right triangle. so you do c^2 - a^2 = b^2

you get 9 for b^2 because 25-16=9 and so you put that in the equation

3 0
3 years ago
Read 2 more answers
In physics, if a moving object has a starting position at so, an initial velocity of vo, and a constant acceleration a, the
emmasim [6.3K]

Answer:

a=\frac{2S -2v_ot-2s_o}{t^2}

Step-by-step explanation:

We have the equation of the position of the object

S = \frac{1}{2}at ^2 + v_ot+s_o

We need to solve the equation for the variable a

S = \frac{1}{2}at ^2 + v_ot+s_o

Subtract s_0 and v_0t on both sides of the equality

S -v_ot-s_o = \frac{1}{2}at ^2 + v_ot+s_o - v_ot- s_o

S -v_ot-s_o = \frac{1}{2}at ^2

multiply by 2 on both sides of equality

2S -2v_ot-2s_o = 2*\frac{1}{2}at ^2

2S -2v_ot-2s_o =at ^2

Divide between t ^ 2 on both sides of the equation

\frac{2S -2v_ot-2s_o}{t^2} =a\frac{t^2}{t^2}

Finally

a=\frac{2S -2v_ot-2s_o}{t^2}

5 0
3 years ago
Which equation is a point slope form equation for line AB ?
kolezko [41]
I think the answer is A
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3 years ago
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5x-3x(26x-7x-28x) systems of equation
labwork [276]

Simplifying
5x + 2(8x + -9) = 3(x + 4) + -5(2x + 7)

Reorder the terms:
5x + 2(-9 + 8x) = 3(x + 4) + -5(2x + 7)
5x + (-9 * 2 + 8x * 2) = 3(x + 4) + -5(2x + 7)
5x + (-18 + 16x) = 3(x + 4) + -5(2x + 7)

Reorder the terms:
-18 + 5x + 16x = 3(x + 4) + -5(2x + 7)

Combine like terms: 5x + 16x = 21x
-18 + 21x = 3(x + 4) + -5(2x + 7)

Reorder the terms:
-18 + 21x = 3(4 + x) + -5(2x + 7)
-18 + 21x = (4 * 3 + x * 3) + -5(2x + 7)
-18 + 21x = (12 + 3x) + -5(2x + 7)

Reorder the terms:
-18 + 21x = 12 + 3x + -5(7 + 2x)
-18 + 21x = 12 + 3x + (7 * -5 + 2x * -5)
-18 + 21x = 12 + 3x + (-35 + -10x)

Reorder the terms:
-18 + 21x = 12 + -35 + 3x + -10x

Combine like terms: 12 + -35 = -23
-18 + 21x = -23 + 3x + -10x

Combine like terms: 3x + -10x = -7x
-18 + 21x = -23 + -7x

Solving
-18 + 21x = -23 + -7x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '7x' to each side of the equation.
-18 + 21x + 7x = -23 + -7x + 7x

Combine like terms: 21x + 7x = 28x
-18 + 28x = -23 + -7x + 7x

Combine like terms: -7x + 7x = 0
-18 + 28x = -23 + 0
-18 + 28x = -23

Add '18' to each side of the equation.
-18 + 18 + 28x = -23 + 18

Combine like terms: -18 + 18 = 0
0 + 28x = -23 + 18
28x = -23 + 18

Combine like terms: -23 + 18 = -5
28x = -5

Divide each side by '28'.
x = -0.1785714286

Simplifying
x = -0.1785714286
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3 years ago
Which of these values represents the longest period of time?
laiz [17]
The answer is D. 6 days 

A. 60 hours is 3600 minutes
B. 60,000 milliseconds is   0.016 minutes
C. 6,000 minutes
D.  6 days is 8640 minutes
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3 years ago
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