All categories

3 years ago

Evaluate 6+4/a+b/3 when a=4 and b=3

Mathematics

1 answer:

vodka [1.7K]3 years ago

6 0

Answer:

Step-by-step explanation:

First, insert the values into the problem. It would look like this: 6 + 4/4 + 3/3

We can then reduce 4/4 into 1 and 3/3 into 1 as well.

Then, you get:

6+1+1 or 6+2 which is equal to 8

-Cheers, Galaxy

You might be interested in

Wyatt plays on the school baseball team. In the last 10 games, Wyatt was

Ksju [112]

Answer:

Step-by-step explanation:

21 hits / 45 at bats

7/15

5 0

2 years ago

the distance from the earth to the sun is about 93,000,000 miles. how many pennies would it take to make a stack that high? can

oksano4ka [1.4K]

Answer:

84,480 Pennies

Step-by-step explanation:

I hope this helps!

7 0

2 years ago

Read 2 more answers

5x - 6 = 3x - 8 what is the answer pls

marishachu [46]

The answer is x = - 1 i am positive

7 0

3 years ago

Read 2 more answers

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

4 0

2 years ago

Describe how to write an equation from a table of values using just the data.

Rama09 [41]

Look at the coordinates and trace those points to a corresponding number on the x or y axis. Then write the first number in the coordinate in the x value and the second number for the y value, repeat the process for the rest of the points.

8 0

3 years ago

Read 2 more answers