Given that the measure of ∠x is 110°, and the measure of ∠y is 59°, find the measure of ∠z.

PLEASE SOMEONE HELP I TRULY DO NOT UNDERSTAND

IceJOKER [234]

Hi there!

$\large\boxed{f^{-1}(x) = \sqrt[3]{\frac{x+4}{9} } }$

$f(x) = 9x^{3} - 4$

Find the inverse by replacing f(x) with y and swapping the x and y variables:

$x = 9y^{3} - 4$

Isolate y by adding 4 to both sides:

$x + 4 = 9y^{3}$

Divide both sides by 9:

$\frac{x+4}{9}= y^{3}$

Take the cube root of both sides:

$y = \sqrt[3]{\frac{x+4}{9} }\\\\f^{-1}(x) = \sqrt[3]{\frac{x+4}{9} }$

7 0

3 years ago

In sunlight, a vertical stick has a height of 5 ft and casts a shadow 3 ft long at the same time that a nearby tree casts a shad

leva [86]

The tree is 25 feet tall. Given the height of the stick and the shadow it cast, the angle formed by the sun and the stick's height can be obtained by taking the Inverse Tangent of 3/5. This is equal to 30.93. This angle is equal to the angle formed by the sun and the tree's height. Using the tangent formula, Tan (30.93)=tree's shadow (15 ft)/ height of the tree, giving the answer 25 feet.

7 0

3 years ago

Read 2 more answers

Craig ran the first part of a race with an average speed of 8 miles per hour and biked the second part of a race with an average

lutik1710 [3]

Let the distance of the first part of the race be x, and that of the second part, 15 - x, then
x/8 + (15 - x)/20 = 1.125
5x + 2(15 - x) = 40 x 1.125
5x + 30 - 2x = 45
3x = 45 - 30 = 15
x = 15/3 = 5

Therefore, the distance of the first part of the race is 5 miles and the time is 5/8 = 0.625 hours or 37.5 minutes
The distance of the second part of the race is 15 - 5 = 10 miles and the time is 1.125 - 0.625 = 0.5 hours or 30 minutes.

4 0

3 years ago

Read 2 more answers

A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}$

By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}$

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

$V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}$ , where $c(0) = 0 \frac{pounds}{gallon}$ .

$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

The solution of this equation is:

$c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})$

The salt concentration after 8 minutes is:

$c(8) = 0.166 \frac{pounds}{gallon}$

The instantaneous amount of salt in the tank is:

$m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds$

3 0

3 years ago

Y=6x+5 is the equation. what is its gradient

Irina18 [472]

Gradient is synonymous to slope.

y = 6x + 5 is an example of a slope-intercept form of a straight-line equationl.

The slope-intercept equation is like this:

y = mx + b

where : m = slope ; b = "y-intercept"

y = 6x + 5

6 is the slope
5 is the y-intercept.

Since slope and gradient are synonymous, 6 is the gradient of the above equation.

8 0

3 years ago