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3 years ago

What is the second step to solve this x/2-5=1

Mathematics

1 answer:

sveta [45]3 years ago

5 0

Answer:

X = -3

Step-by-step explanation:

X/2-5 = 1

X/-3 = 1

Multiply both sides by -3 to isolate x

X = -3

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Find the slope of the line through the given pair of points.<br> (0,16), (11,6)

Lelechka [254]

Answer:

m = $-\frac{10}{11}$

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3 years ago

Read 2 more answers

Find the dimensions of the rectangle of maximum area that can be formed from a 210-in. piece of wire. (Use decimal notation. Giv

Monica [59]

Answer:

52.500 by 52.500 inches

Step-by-step explanation:

The rectangle with maximum area will be a square. Its side length will be 1/4 the perimeter, so is 210/4 = 52.5 inches.

The figure is a 52.500 inch square. The interval of optimization is <em>closed</em>.

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Side lengths are restricted to the interval 0 to 105 inches.

Any n-sided polygon with a given perimeter will have its maximum area when the polygon is regular. A regular 4-gon is a square.

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3 years ago

Consider the equation below. (If an answer does not exist, enter DNE.)f(x) = x2 − x − ln(x)

Morgarella [4.7K]

Answer:

a) decreasing from (0,2), increasing from (2,∞ )

b) local minimum in x=2 . there is no maximum or minimum value

c) DNE. there is no inflexion point

Step-by-step explanation:

f(x) = x² - x - ln (x)

since ln(x) is defined for positive values only x must be greater than 0 (x>0)

also we will need the first derivative and the second derivative with respecto to x

f(x) = x² - x - ln (x)

df/dx (x) = 2x - 1 - 1/x

d²f /(dx)² (x) = 2 + 1/x²

a) to find the increasing and decreasing intervals we will need to evaluate the rate of change (df/dx) :

df/dx = 0 when 2x - 1 - 1/x = 0 → 2x² - x - 1 = 0 → x = (1±√(1+8))/2 = (1 ± 3)/2

→ x1 = 2 , x2 = -1 (discarded because x2<0)

therefore since 2x increases and 1/x decreases with increasing x

for x > 2 , df/dx is positive and thus f increases with increasing x

for 0<x< 2, df/dx is negative and thus f decreases with increasing x

b) since f increases with increasing x for x> 2 and decreases with increasing x for, 0<x< 2 , f should be a minimum value.

we can verify it with the second derivative

d²f /(dx)² (x) =2 + 1/x² → for x >0 , d²f /(dx)² is always >0 therefore

d²f /(dx)² (x1) > 0 and df/dx (x1) =0 → thus f(x) is a local minimum of x

there are no maximum values since for x → ∞ , f(x) → ∞ and for x→ 0 → f(x) → -∞ (because of the ln(x) function)

c) there are no inflexion points since d²f /(dx)² (x1) is always greater than 0 for x>0

4 0

3 years ago

F(x)=-2(x-1)^2+2 a) standard form b) graph

OverLord2011 [107]

Answer:

a) f(x) = -2x² + 4x

Explanation:

To find the standard form we need to solve the equation as:

$\begin{gathered} f(x)=-2(x-1)^2+2 \\ f(x)=-2(x^2-2x+1)+2 \\ f(x)=-2x^2-2(-2x)-2(1)+2 \\ f(x)=-2x^2+4x-2+2 \\ f(x)=-2x^2+4x \end{gathered}$

So, the standard form is f(x) = -2x² + 4x

To make the graph we will use the initial form because when the equation is written like f(x) = a(x-h)²+k, the coordinate (h,k) is the vertex of the parabola

So, in this case, the vertex of the parabola is the point (1, 2)

On the other hand, we can find a point in the graph. For example, if x is equal to 0, then f(x) is equal to:

f(x) = -2(x-1)² + 2

f(0) = -2(0-1)² + 2

f(0) = -2(-1)² + 2

f(0) = - 2 + 2

f(0) = 0

So, the parabola passes through the point (0,0) and has a vertex in the point (1, 2). Then, the graph is:

4 0

1 year ago

Barbra brought 5 amusement park tickets at a cost for $30 if she bought 7 tickets,how much would it cost

castortr0y [4]

5 tickets for $30
this means one ticket is $6 as 30÷5=6
7 multiplied by 6= $42
7 tickets would cost $42

4 0

3 years ago