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3 years ago

Add the following - 4/9,7/12and - 3/8

Mathematics

1 answer:

bulgar [2K]3 years ago

8 0

Answer:

Step-by-step explanation:

$- \frac{4}{9} + \frac{7}{12} + ( - \frac{ 3}{8} )$

When there is a (+) in front of an expression in parentheses, the expression remains the same:

$- \frac{4}{9} + \frac{7}{12} - \frac{3}{8}$

$\frac{ - 4 \times 8 + 7 \times 6 - 3 \times 9}{72}$

Calculate the sum of difference

$\frac{ - 32 + 42 - 27}{72}$

$\frac{10 - 27}{72}$

$- \frac{17}{72}$

Hope this helps..

Good luck on your assignment...

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4. Simplify the expression by combining like terms. 8a+6b+5a+2b *

marissa [1.9K]

Answer:

Step-by-step explanation:

Add the like terms, the a's by themselves and b's by themselves.

8a+5a=13a

6b+2b=8b

so 13a+8b

7 0

3 years ago

A trough is 9 feet long, and its cross section is in the shape of an isosceles right triangle with hypotenuse 2 feet, as shown a

Andrei [34K]

Answer:

1/6

Step-by-step explanation:

Given:

Length of the trough: 9 ft

=> The volume of the trough: V = $\frac{9}{2}$ * (b * h) (1)

An isosceles right triangle with hypotenuse 2 feet

=> the other two sides of the triangle is:

= tan(45 degrees) = h/(b/2)

<=> b = 2h substitute in (1), we have:

V = $\frac{9}{2}$ *(2h *h) = 9 $h^{2}$

Take derivative of volume with respect to time to find equation for rate of filling the trough

dV/dt = 2 * 9 *h dh/dt = 18h dh/dt

<=> dh/dt = dV/dt /(18h)

As we know that, dV/dt = 2

So, dh/dt = 2 / 18h = 1/9h

<=> V = t * rate = 2 * 2 = 4

But V = 9 $h^{2}$

<=> 9 $h^{2}$ = 4

<=> h = 2/3

The rate is the height h feet of the water in the trough changing 2 minutes after the water begins to flow:

dh/dt = 1/(9h) = 1/(9 * 2/3) = 1/6

8 0

3 years ago

Suppose u and v are functions of x that are differentiable at x=0 and that u(0)=4, u′(0)=7, v(0)=2 and v′(0)=1. Find the values

Leona [35]

Answer with Step-by-step explanation:

We are given that u and v are functions of x and both are differentiable at x=0

$u(0)=4,u'(0)=7,v(0)=2,v'(0)=1$

a.We have to find the values of $\frac{d(uv)}{dx}$

$\frac{d(u\cdot v)}{dx}=u'v+uv'$

Using this formula

Then , we get

$[\frac{d(uv)}{dx}]_{x=0}=u'(0)v(0)+u(0)v'(0)=7(2)+4(1)=14+4=18$

$[\frac{d(uv)}{dx}]_{x=0}=18$

b. $\frac{d(u/v)}{dx}=\frac{u'v-uv'}{v^2}$

$[\frac{d(u/v)}{dx}]_{x=0}=\frac{u'(0)v(0)-u(0)v'(0)}{v^2(0)}=\frac{7(2)-4(1)}{2^2}=\frac{14-4}{4}=\frac{10}{4}=\frac{5}{2}$

$[\frac{d(u/v)}{dx}]_{x=0}=\frac{5}{2}$

$[\frac{d(v/u)}{dx}]_{x=0}=\frac{v'(0)u(0)-v(0)u'(0)}{u^2(0)}=\frac{1(4)-7(2)}{4^2}$

$[\frac{d(v/u)}{dx}]_{x=0}=\frac{-10}{16}=\frac{-5}{8}$

d. $\frac{d(-6v-9u)}{dx}=-6v'-9u'$

$[\frac{d(-6v-9u)}{dx}]_{x=0}=-6v'(0)-9u'(0)=-6(1)-9(7)=-6-63=-69$

$[\frac{d(-6v-9u)}{dx}]_{x=0}=-69$

3 0

3 years ago

A local hamburger shop sold a combined total of 578 hamburgers and cheeseburgers on Tuesday. There were 72 fewer cheeseburgers s

andreev551 [17]

Let h be hamburgers and c be cheeseburgers.
We know that $h + c = 578$
We also know that $c + 72 = h$
This gives us a system of equations we can solve. My preferred method to solve systems is substitution, where we solve one equation for one of the variables, then substitute that solution in the other equation, reducing it to a single variable equation. One of the equations already equals h, so we can go straight into the sub part.
$h + c = 578$
$(c + 72) + c = 578$
$c2 = 506$
$c = 253$
Finally, we go back to the other equation and solve for h.
$c + 72 = h$
$253 + 72 = h$
$325 = h$
So, total there were 253 cheeseburgers sold, and 325 hamburgers sold.

6 0

4 years ago

Read 2 more answers

What is the value of 5a when a=7 ?

Mariulka [41]

Answer:

...

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7 0

3 years ago

Read 2 more answers

Add the following - 4/9,7/12and - 3/8​

Add the following - 4/9,7/12and - 3/8