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3 years ago

Solve. -7 2/3+(-5 1/2)+8 3/4

Mathematics

1 answer:

choli [55]3 years ago

4 0

Answer:

I'm pretty sure it's 21 11/12. Don't get mad if I'm wrong

Step-by-step explanation:

-7 2/3 - 5 1/2 + 8 3/4

= -7 8/12 - 5 6/12 + 8 9/12

12 14/12 + 8 9/12

20 23/12

= 21 11/12

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FIND THE AREA (LEAVE EXPLANATION)

VladimirAG [237]

Answer:

20 yd²

Step-by-step explanation:

lets do the big one

4 x (6-2) = 4 x 4 = 16 yd²

2 x (4-2) = 2x2 = 4 yd²

16 yd² + 4 yd² = 20 yd²

Another way

Find the area of the figure if it was a rectangle

4 x 6 = 24 yd²

then subtract 2 x 2 = 24 yd² - 4 yd² = 20 yd²

4 0

3 years ago

Read 2 more answers

Can someone PLEASE explain to me how to find the rule of a function table. I understand how find the rule of easy function table

goldfiish [28.3K]

Answer:

y=2x-46

Step-by-step explanation:

Select any two coordinate points. I'm picking (1,-44) and (2, -42).

x1 and y1 are the first coordinates. x2 and y2 are the second coordinates. The formula for finding the slope of a linear fuction is (y2-y1)/(x2-x1). Substiute the numbers into this formula:

-42-(-44)/2-1

=-42+44/1

This is the slope. In y=mx+b, plug in your newly found slope. Next, to find the y-intercept, plug any pair of coordinate points into your incomplete formula y=2x+b. I'm going to use (1,-44) again.

-44=2(1)+b

-44=2+b

b=-46

Now, plug this into your formula, and you'll get y=2x-46.

4 0

3 years ago

Read 2 more answers

Given that ABCD is a rhombus, what is the value of x?

Maslowich

Answer:

Step-by-step explanation:

We know that diagonals bisect angles of a rhombus, meaning that they split the angles of two equal parts, that opposite angles are equal, and that the angles of a quadrilateral, such as a rhombus, add up to 360 degrees.

Looking at the drawing I uploaded, and taking into account that diagonals bisect angles of a rhombus, we can say that angle x = y and z = 5x-18

Next, given that opposite angles are equal, we can say that angle B = angle D and angle A = angle C. Therefore,

A+B+C+D= 360 (as the angles of a quadrilateral add up to 360)

A + B + A + B = 360 (plugging in A for C and B for D)

2 ( A + B) = 360

A + B = 180

x + y + z + 5x - 18 = 180 (plugging x+y in for A, and z+5x-18 in for B, as x and y make up A and z and 5x-18 make up B)

x+x+5x-18+5x-18 = 180 (because x=y and z=5x-18)

2(x+5x-18) = 180

divide both sides by 2

x+5x-18=90

6x-18=90

add 18 to both sides

6x=108

divide both sides by 6

x = 18

4 0

3 years ago

For the parallelogram, find the value of the variables. Show your work.

coldgirl [10]

Of course you already know everything parallel sides are equals
therefore the value of each variables will be as following :
left side = right side
3y-6=24
3y=24+6
3y=30
y=30/3
y=10
-----------------------------------------------
5x+2=12
5x=12-2
5x=10
x=10/5
x=2

6 0

3 years ago

The concentration C of certain drug in a patient's bloodstream t hours after injection is given by

frozen [14]

Answer:

a) The horizontal asymptote of $C(t)$ is $c = 0$ .

b) When t increases, both the numerator and denominator increases, but given that the grade of the polynomial of the denominator is greater than the grade of the polynomial of the numerator, then the concentration of the drug converges to zero when time diverges to the infinity. There is a monotonous decrease behavior.

c) The time at which the concentration is highest is approximately 1.291 hours after injection.

Step-by-step explanation:

a) The horizontal asymptote of $C(t)$ is the horizontal line, to which the function converges when $t$ diverges to the infinity. That is:

$c = \lim _{t\to +\infty} \frac{t}{3\cdot t^{2}+5}$ (1)

$c = \lim_{t\to +\infty}\left(\frac{t}{3\cdot t^{2}+5} \right)\cdot \left(\frac{t^{2}}{t^{2}} \right)$

$c = \lim_{t\to +\infty}\frac{\frac{t}{t^{2}} }{\frac{3\cdot t^{2}+5}{t^{2}} }$

$c = \lim_{t\to +\infty} \frac{\frac{1}{t} }{3+\frac{5}{t^{2}} }$

$c = \frac{\lim_{t\to +\infty}\frac{1}{t} }{\lim_{t\to +\infty}3+\lim_{t\to +\infty}\frac{5}{t^{2}} }$

$c = \frac{0}{3+0}$

$c = 0$

The horizontal asymptote of $C(t)$ is $c = 0$ .

c) From Calculus we understand that maximum concentration can be found by means of the First and Second Derivative Tests.

First Derivative Test

The first derivative of the function is:

$C'(t) = \frac{(3\cdot t^{2}+5)-t\cdot (6\cdot t)}{(3\cdot t^{2}+5)^{2}}$

$C'(t) = \frac{1}{3\cdot t^{2}+5}-\frac{6\cdot t^{2}}{(3\cdot t^{2}+5)^{2}}$

$C'(t) = \frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right)$

Now we equalize the expression to zero:

$\frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right) = 0$

$1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} = 0$

$\frac{3\cdot t^{2}+5-6\cdot t^{2}}{3\cdot t^{2}+5} = 0$

$5-3\cdot t^{2} = 0$

$t = \sqrt{\frac{5}{3} }\,h$

$t \approx 1.291\,h$

The critical point occurs approximately at 1.291 hours after injection.

Second Derivative Test

The second derivative of the function is:

$C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}-\frac{(12\cdot t)\cdot (3\cdot t^{2}+5)^{2}-2\cdot (3\cdot t^{2}+5)\cdot (6\cdot t)\cdot (6\cdot t^{2})}{(3\cdot t^{2}+5)^{4}}$

$C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}- \frac{12\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}$

$C''(t) = -\frac{18\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}$

If we know that $t \approx 1.291\,h$ , then the value of the second derivative is:

$C''(1.291\,h) = -0.077$

Which means that the critical point is an absolute maximum.

The time at which the concentration is highest is approximately 1.291 hours after injection.

5 0

3 years ago