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const2013 [10]
3 years ago
5

Which expression is equivalent to 3m+1-m?

Mathematics
2 answers:
Lisa [10]3 years ago
6 0

Answer:

option  (a) is correct.

2 + m - 1 + m is an equivalent expression  to the given expression  3m+1-m

Step-by-step explanation:

 Given expression  3m+1-m

We have to choose an equivalent expression from given options.

Equivalent expression are those expression that looks different but are same.

Like 4+2 = 6  and 3+ 3 = 6

Both have same value but looks differently.

Like terms are term having same variable with same degree.

Consider the given expression  3m+1-m

Simplify by adding like terms,

3m + 1 - m = (3-1) m + 1

Thus, (3-1) m + 1  = 2m + 1

Also 2m + 1 can be written as m + m + 2 - 1

Thus, option  (a) is correct.

Umnica [9.8K]3 years ago
5 0

Answer:

a

Step-by-step explanation:

i took the test. i hope this helps !!! :D

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The width of a rectangle is 30% of its length what is the area of the rectangle. when the length is 40ft.
zaharov [31]

Answer:

480ft squared

Step-by-step explanation:

length= 40*0.30= 12

area= length×width

area=40×12

40×12= 480

7 0
3 years ago
A grasshopper jumps straight up from the ground with an initial vertical velocity of 8 feet per second. After how many seconds w
vlabodo [156]

Answer:

Problem 1 : After 0.25 s

Problem 2 : The squirrel doesn't reach the ground before the nut

Step-by-step explanation:

Problem 1

This is a standard physics problem

The equation that defines the movement of the grasshopper given the initial velocity, and the acceleration is the following

Distance =  Do + Vo*t + 0.5*a*t^2

Where

Do is the initial distance.

(Do = 0, because the grasshopper is on the ground and we choose the reference system there)

Vo is the initial velocity = 8 feet/sec

a is the acceleration that is exerted on the body (in this case a = g = -9.8 m/s^2 = - 32.174 feet per second per second.)

t is the time

We substitute in the equation, and solve for t (time)

Distance =  Do + Vo*t + 0.5*a*t^2

(1 foot)=  (0)+ (8 feet/s)*t + 0.5*(-32.174 feet/s^2)*t^2

Solving this equation, we get that

t = 0.25 s

Problem 2

This problem is similar to the previous one

We apply the same formula

D =  Do + Vo*t + 0.5*a*t^2

But in this case, we are going in the opposite direction, and we have an initial distance.

Do = 27 feet

Vo is the initial velocity of the nut = -6 feet/s

D = 0  (Ground)

We substitute in the equation, and solve for t (time)

(0) =  (27) + (-6 feet/s)*t + 0.5*(-32 feet/s^2)*t^2

t = 1.125 s  <  2s

The nut reaches ground in 1.125 s, so the squirrel doesn't reach the ground before the nut

4 0
3 years ago
How do I find how many times a line intersects?
MAXImum [283]

Answer:

you should count how many time two or more line touches each other

5 0
3 years ago
Read 2 more answers
Quarters make up what part of the group of coins ? Write the answer as fraction and word firm
aleksandr82 [10.1K]
Quarts = 25 there are 100 in a dollar so you have 25 out of 100

25/100<span />
3 0
3 years ago
Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre
Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

(c) Inflection point: (0.56,-0.06)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = \frac{d}{dx}[x^{4}ln(x)]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = 4x^{3}ln(x) + x_{4}.\frac{1}{x}

f'(x) = 4x^{3}ln(x) + x^{3}

f'(x) = x^{3}[4ln(x) + 1]

Now, find the critical points: f'(x) = 0

x^{3}[4ln(x) + 1] = 0

x^{3} = 0

x = 0

and

4ln(x) + 1 = 0

ln(x) = \frac{-1}{4}

x = e^{\frac{-1}{4} }

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval                 x-value                      f'(x)                       result

0<x<0.78                 0.5                 f'(0.5) = -0.22            decreasing

x>0.78                       1                         f'(1) = 1                  increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

  • Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
  • After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = x^{4}ln(x)

f(0.78) = 0.78^{4}ln(0.78)

f(0.78) = - 0.092

The point of <u>minimum</u> is (0.78, - 0.092)

(c) To determine the inflection point (IP), calculate the second derivative of the function and solve for x:

f"(x) = \frac{d^{2}}{dx^{2}} [x^{3}[4ln(x) + 1]]

f"(x) = 3x^{2}[4ln(x) + 1] + 4x^{2}

f"(x) = x^{2}[12ln(x) + 7]

x^{2}[12ln(x) + 7] = 0

x^{2} = 0\\x = 0

and

12ln(x) + 7 = 0\\ln(x) = \frac{-7}{12} \\x = e^{\frac{-7}{12} }\\x = 0.56

Substituing x in the function:

f(x) = x^{4}ln(x)

f(0.56) = 0.56^{4} ln(0.56)

f(0.56) = - 0.06

The <u>inflection point</u> will be: (0.56, - 0.06)

In a function, the concave is down when f"(x) < 0 and up when f"(x) > 0, adn knowing that the critical points for that derivative are 0 and 0.56:

f"(x) =  x^{2}[12ln(x) + 7]

f"(0.1) = 0.1^{2}[12ln(0.1)+7]

f"(0.1) = - 0.21, i.e. <u>Concave</u> is <u>DOWN.</u>

f"(0.7) = 0.7^{2}[12ln(0.7)+7]

f"(0.7) = + 1.33, i.e. <u>Concave</u> is <u>UP.</u>

4 0
3 years ago
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