What is the solution to the system of equations? y = A system of equations. Y equals StartFraction one-third EndFraction minus 1

Respuesta:

El total de la factura de Isaac es $84.53

Explicación detallada:

$37.00 + $42.00 = $79.00 < b (factura)

b + t (impuesto sobre ventas) = c (costo total)

(7%/100 = 0.07)

(0.07 * 79 = 5.53)

(t = $5.53)

$79.00 + $5.53 = $84.53

¡Buena suerte!

Answer:

Find below:

Step-by-step explanation:

To determine this, we will either calculate the total cost of acquiring 40 bouquets at $2.50 each or find the single bouquet’s cost at $120.

Cost of one in pack of 40 priced at $120.

120 divided by 40 equals $3

Now, we notice that $3>$2.50

This indicates Kendra has made an error by purchasing the 40 bouquet pack at $120

Hope this helps.

Good Luck

Response:

• Solution for x is 1

Detailed explanation:

Line PU acts as a transversal intersecting parallel lines QR and ST

The angles PRQ and UST sum to 180 degrees

• 15(x+2) = 180 - 135
• 15x + 30 = 45
• 15x = 15
• x = 1

It exhibits a zero slope.

In the case of a horizontal line, the slope remains constant at 0 since there is no variation.

Answer:

Alright, we can express the baseball's motion with an equation like:

h(x) = a*x^2 + b*x + c

Here, x denotes time, while h(x) indicates height.

Let’s construct this:

The acceleration is:

a(t) = a

For velocity, integrating over time results in:

v(x) = a*x + v0

Where v0 signifies the initial vertical velocity.

Subsequently, we can determine position or height by integrating once more:

h(x) = a*x^2 + v0*x + h0

Here, h0 is the initial height.

<pthus our="" equation="" is:="">

h(x) = a*x^2 + v0*x + h0.

<pexamining the="" table:="">

When x = 0s, h(0s) = 6ft

<pthus:>

h(0s) = a*0s^2 + v0*0s + h0 = 6ft

            h0 = 6ft.

It’s also noted that:

h(2s) = h(4s)

<pthe symmetry="" of="" the="" quadratic="" function="" implies="" that="" axis="" lies="" between="" and="" located="" at="" x="3s.&lt;/p"><pin a="" standard="" quadratic="" function:="">

a*x^2 + b*x + c

The symmetry line is given by:

x = -b/2a

<pin this="" instance:="">

b = v0

a = a

<ptherefore we="" derive:="">

3s  = -v0/(2*a)

v0 = -3s*(2a)

<phaving gathered="" all="" necessary="" data="" for="" our="" equation="" we="" can="" express="" it="" as:="">

h(x) = a*x^2 - 6s*a*x + 6ft

<pnext focusing="" on="" just="" one="" variable="" we="" know="" that="" at="" x="2s," h=""><pso:>

h(2s) = 22ft = a*(2s)^2 - 6s*a*2s + 6ft

<pthus our="" resulting="" equation="" reads:="">

h(x) = (-2ft/s^2)*x^2 + (12ft/s)*x + 6ft

b) The height after 5 seconds is expressed as:

h(5s) =  (-2ft/s^2)*(5s)^2 + (12ft/s)*5s + 6ft = 16ft

</pthus></p></pso:></pnext></phaving></ptherefore></pin></pin></pthe></pthus:></pexamining></pthus>