What type of System if the two linear equations have only one solution?

Can somebody help me solve 8^4v + 5 = 31 ?

bezimeni [28]

$8^{4v}+5=31$
minus 5 both sides
$8^{4v}=26$
take the log₈ of both sides
4v=log₈26
divide both sides by 4
v=(1/4)log₈28

if yo use calculator
go back
$8^{4v}=26$
take ln of both sides
4vln8=ln26
divide both sides by (4ln8)
v=(ln28)/(4ln8)
evaluate
v≈0.40061291017147

3 0

2 years ago

Please please please help me with this answer

Rasek [7]

x = number of packages he bought

y = number of popsicles in each package

Here are the two equations:

xy = 54

x = 3 + y

Using the substitution method, you can solve them to find x and y.

(3 + y)y = 54,

3y + y^2 = 54

y^2 + 3y - 54 = 0

Solve the quadriatic equation:

(y +9)(y - 6) = 0

y = -9, 6.

Plug y into: xy = 54

x(-9) = 54, x = -6.

x(6) = 54, x = 9.

So he bought 9 packages of popsicles, bc you can't bye -6 packages.

3 0

3 years ago

To solve y and x, I am adding a picture

Viktor [21]

Answer:

$x =\sqrt 5$

$y = \sqrt{5$

Step-by-step explanation:

Given

The attached triangle

Required

Solve for x

Considering angle 45 degrees, we have:

$\cos(45) = \frac{y}{\sqrt{10}}$ --- cosine formula i.e. adj/hyp

Solve for y

$y = \sqrt{10} * \cos(45)$

In radical form, we have:

$y = \sqrt{10} * \frac{1}{\sqrt 2}$

$y = \sqrt{10/2}$

$y = \sqrt{5$

To solve for x, we make use of Pythagoras theorem

$x^2 + y^2 = (\sqrt{10})^2$

$x^2 + y^2 =10$

Substitute for y

$x^2 + (\sqrt 5)^2 =10$

$x^2 + 5 =10$

Collect like terms

$x^2 =10-5$

$x^2= 5$

Solve for x

$x =\sqrt 5$

4 0

3 years ago

The total cost (in hundreds of dollars) to produce x units of a product is c(x) = (3x-2) / (8x+1), find the average cost for eac

olya-2409 [2.1K]

Answer:

a) $\frac{74}{10025}$

b) $\frac{3x-2}{x(8x+1)}$

c) $\frac{-24x^2+32x-2}{(8x^2+x)^2}$

Step-by-step explanation:

For total cost function $c(x)$ , average cost is given by $\frac{c(x)}{x}$ i.e., total cost divided by number of units produced.

Marginal average cost function refers to derivative of the average cost function i.e., $\left ( \frac{c(x)}{x} \right )'$

Given: $c(x)=\frac{3x-2}{8x+1}$

Average cost = $\frac{c(x)}{x}=\frac{3x-2}{x(8x+1)}$

a)

At x = 50 units,

$\frac{c(50)}{50}=\frac{150-2}{50(400+1)}=\frac{148}{50(401)}=\frac{74}{10025}$

b)

Average cost = $\frac{c(x)}{x}=\frac{3x-2}{x(8x+1)}$

c)

Marginal average cost:

Differentiate average cost with respect to $x$

Take $f=3x-2\,,\,g=8x^2+x$

using quotient rule, $\left ( \frac{f}{g} \right )'=\frac{f'g-fg'}{g^2}$

Therefore,

$\left ( \frac{c(x)}{x} \right )'=\left ( \frac{3x-2}{8x^2+x} \right )'\\=\left ( \frac{3(8x^2+x)-(16x+1)(3x-2)}{(8x^2+x)^2} \right )\\=\frac{24x^2+3x-48x^2-3x+32x+2}{(8x^2+x)^2}\\=\frac{-24x^2+32x-2}{(8x^2+x)^2}$

3 0

3 years ago

I need help on 16 please

jonny [76]

I believe it would be 65

5 0

3 years ago