Solve for x. −5x−4(x−6)=−3

Find the mean of the following number set: 89, 76, 85, 76, 77, 84

klio [65]

Answer:

81.167~

Step-by-step explanation:

You can find the mean (or average) of a set of numbers by adding them together and dividing the sum by the amount of values there are.

In this case, there are 6 values.

The total sum of the six values is 487.

To receive the mean, divide 487 by 6.

6 0

3 years ago

Read 2 more answers

At the city museum, child admission is $6.00 and adult admission is $9.20. On Tuesday, twice as many adult tickets as child

jekas [21]

Answer:

38

Step-by-step explanation:

3 0

3 years ago

Marcos had 15 coins in nickels and quarters. He had 3 more quarters than nickels. He wrote a system of equations to represent th

hodyreva [135]

Answer:

There are 6 coins in nickels and 9 coins in quarters.

Step-by-step explanation:

Now, to find the solution.

According to question:

The number of nickels = $x$ .

The number of quarters = $y.$

So, the total number of coins:

$x+y=15.$

He had 3 more quarters than nickels.

Thus,

$y=x+3$ .....(1)

Now, to get the solution:

$x+y=15$

Substituting the equation (1):

$x+(x+3)=15$

$x+x+3=15$

$2x+3=15$

Subtracting both sides by 3 we get:

$2x=12$

Dividing both sides by 2 we get:

$x=6.$

The number of nickels = 6.

Now, substituting the value of $x$ in equation (1):

$y=x+3\\\\y=6+3\\\\y=9.$

The number of quarters = 9.

Thus, there are 6 coins in nickels and 9 coins in quarters.

6 0

3 years ago

PLZ HELP! VERY URGENT! A line passes through (6,3), (8,4), and (n,-2). Find the value of n.

Phantasy [73]

We have a line passing through 3 points. We can use the first
two points to find the equation of the line.
(6,3), (8,4)

slope m = (4-3)/(8-6) = 1/2

We can use the point slope form y-y1 = m(x-x1) or
the slope intercept form y = mx + b
to find the equation of the line.

Let's take the point (6,3) and use y = mx + b to find b
y = 3, x = 6, m = 1/2

3 = (1/2)(6) + b
3 = 3 + b
0 = b

The equation of our line is y = (1/2)x

We have a line of equation y = (1/2)x going through point (n,-2)

Plugging in we have: -2 = (1/2)n
2(-2) = n
-4 = n

Your answer is n = -4

NOTE: looking at the 3 points as given initially: (6,3), (8,4), (n,-2)
We can see that 3 = (1/2)6 and 4 = (1/2)8 so -2 = (1/2)n makes sense

Hope this helps you :)

5 0

3 years ago

Read 2 more answers

Let R be the region in the first quadrant of the xy-plane bounded by the hyperbolas xyequals1, xyequals9, and the lines yequa

Tema [17]

Answer:

The area can be written as

$\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = 0.2274$

And the value of it is approximately 1.8117

Step-by-step explanation:

x = u/v

y = uv

Lets analyze the lines bordering R replacing x and y by their respective expressions with u and v.

x*y = u/v * uv = u², therefore, x*y = 1 when u² = 1. Also x*y = 9 if and only if u² = 9

x=y only if u/v = uv, And that only holds if u = 0 or 1/v = v, and 1/v = v if and only if v² = 1. Similarly y = 4x if and only if 4u/v = uv if and only if v² = 4

Therefore, u² should range between 1 and 9 and v² ranges between 1 and 4. This means that u is between 1 and 3 and v is between 1 and 2 (we are not taking negative values).

Lets compute the partial derivates of x and y over u and v

$x_u = 1/v$

$x_v = u*ln(v)$

$y_u = v$

$y_v = u$

Therefore, the Jacobian matrix is

$\left[\begin{array}{ccc}\frac{1}{v}&u \, ln(v)\\v&u\end{array}\right]$

and its determinant is u/v - uv * ln(v) = u * (1/v - v ln(v))

In order to compute the integral, we can find primitives for u and (1/v-v ln(v)) (which can be separated in 1/v and -vln(v) ). For u it is u²/2. For 1/v it is ln(v), and for -vln(v) , we can solve it by using integration by parts:

$\int -v \, ln(v) \, dv = - (\frac{v^2 \, ln(v)}{2} - \int \frac{v^2}{2v} \, dv) = \frac{v^2}{4} - \frac{v^2 \, ln(v)}{2}$

Therefore,

$\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = \int\limits_1^2 (\frac{1}{v} - v \, ln(v) ) (\frac{u^2}{2}\, |_{u=1}^{u=3}) \, dv= \\4* \int\limits_1^2 (\frac{1}{v} - v\,ln(v)) \, dv = 4*(ln(v) + \frac{v^2}{4} - \frac{v^2\,ln(v)}{2} \, |_{v=1}^{v=2}) = 0.2274$

4 0

3 years ago