All categories

3 years ago

What is (-1/2) - (-2/5) ? help please .

Mathematics

2 answers:

Mashutka [201]3 years ago

8 0

Answer:

-1/10

Step-by-step explanation:

Valentin [98]3 years ago

8 0

Answer:

-1/10

Step-by-step explanation:

First you need a common demonator of 10

-1/2 --> -5/10

and -2/5 --> -4/10

using this the equation is:

-5/10 - (-4/10) which equals

-5/10 + 4/10= -1/10

You might be interested in

According to a survey, 15% of city workers take the bus to work. Donatella randomly surveys 10 workers. What is the probability

Andru [333]

Answer:0.001

x=the number of workers taking the bus to work

p= probability of success =(15/100) = 0.15

q= probability of failure =1- p = 0.85

P(X=6) = 10C6(0.15)^6(0.85)^4

= 0.001

4 0

3 years ago

Read 2 more answers

PLEASE HELP!! BE AUICK AND EXPLAIN!Solve the system by graphing!

zubka84 [21]

The answer to the question is (2,-2) d

4 0

3 years ago

The total cost c a painter charges to paint a house depends on the number h of hours it takes to paint the house. This situation

spin [16.1K]

H = 30.25 hours

c = 15(30.25) = 453.75
c = 453.75 + 245
c = 698.75

4 0

3 years ago

Simplify completely quantity 2 x squared plus 16 x plus 30 all over quantity 5 x squared plus 13 x minus 6.

Ludmilka [50]

(2x^2 + 16x + 30) / (5x^2 + 13x -6)
Start by factoring the numerator and denominator.
Cancel any like factors.

2(x^2 + 8x + 15)
2(x + 5)(x + 3) / (5x -2)(x + 3
2(x + 5) / (5x - 2)
The third choice

8 0

3 years ago

Read 2 more answers

Can someone please help me with this

Mama L [17]

Answer:

27.

Equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1: 400 feet

Part 2: Velocity is +96 feet/second (or 96 feet/second UPWARD) & Speed is 96 feet per second

Part 3: acceleration at any time t is -32 feet/second squared

Part 4: t = 10 seconds

29.

Part 1: Average Rate of Change = -15

Part 2: The instantaneous rate of change at x = 2 is -8 & at x = 3 is -23

Step-by-step explanation:

27.

First of all, the equation of tangent line is given by:

$y-y_1=m(x-x_1)$

Where m is the slope, or the derivative of the function

Now,

If we take a point x, the corresponding y point would be $x^2-3x+5$ , so the point would be $(x,x^2-3x+5)$

Also, the derivative is:

$f(x)=x^2-3x+5\\f'(x)=2x-3$

Hence, we can equate the DERIVATIVE (slope) and the slope expression through the point given (0,1) and the point we found $(x,x^2-3x+5)$

The slope is $\frac{y_2-y_1}{x_2-x_1}$

So we have:

$\frac{x^2-3x+5-1}{x-0}\\=\frac{x^2-3x+4}{x}$

Now, we equate:

$2x-3=\frac{x^2-3x+4}{x}$

We need to solve this for x. Shown below:

$2x-3=\frac{x^2-3x+4}{x}\\x(2x-3)=x^2-3x+4\\2x^2-3x=x^2-3x+4\\x^2=4\\x=-2,2$

So, this is the x values of the point of tangency. We evaluate the derivative at these 2 points, respectively.

$f'(x)=2x-3\\f'(-2)=2(-2)-3=-7\\f'(2)=2(2)-3=1$

Now, we find 2 equations of tangent lines through the point (0,1) and with respective slopes of -7 and 1. Shown below:

$y-y_1=m(x-x_1)\\y-1=-7(x-0)\\y-1=-7x\\y=-7x+1$

and

$y-y_1=m(x-x_1)\\y-1=1(x-0)\\y-1=x\\y=x+1$

So equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1:

The highest point is basically the maximum value of the position function. To get maximum, we differentiate and set it equal to 0. Let's do this:

$s(t)=160t-16t^2\\s'(t)=160-32t\\s'(t)=0\\160-32t=0\\32t=160\\t=\frac{160}{32}\\t=5$

So, at t = 5, it reaches max height. We plug in t = 5 into position equation to find max height:

$s(t)=160t-16t^2\\s(5)=160(5)-16(5^2)\\=400$

max height = 400 feet

Part 2:

Velocity is speed, but with direction.

We also know the position function differentiated, is the velocity function.

Let's first find time(s) when position is at 256 feet. So we set position function to 256 and find t:

$s(t)=160t-16t^2\\256=160t-16t^2\\16t^2-160t+256=0\\t^2-10t+16=0\\(t-2)(t-8)=0\\t=2,8$

At t = 2, the velocity is:

$s'(t)=v(t)=160-32t\\v(2)=160-32(2)\\v(2)=96$

It is going UPWARD at this point, so the velocity is +96 feet/second or 96 feet/second going UPWARD

The corresponding speed (without +, -, direction) is simply 96 feet/second

Part 3:

We know the acceleration is the differentiation of the velocity function. let's find it:

$v(t)=160-32t\\v'(t)=a(t)=-32$

hence, the acceleration at any time t is -32 feet/second squared

Part 4:

The rock hits the ground when the position is 0 (at ground). So we equate the position function, s(t), to 0 and find time when it hits the ground. Shown below:

$s(t)=160t-16t^2\\0=160t-16t^2\\16t^2-160t=0\\16t(t-10)=0\\t=0,10$