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Veronika [31]
3 years ago
6

Nico dropped the brand new bottle of dog shampoo and 3 ounce spilled out. He measured 1 1/4 ounces each time he shampoos his dog

and the bottle of shampoo originally contained 30 1/2 ounces. What is an equation and value of x the number of times Nico can bathe his dog using this one bottle
Mathematics
2 answers:
GalinKa [24]3 years ago
7 0

Answer:

equation ( 1.25x = 27.5 ) and x = 22.

Step-by-step explanation:

Shampoo bottle contains shampoo = 30\frac{1}{2} ounces = 30.5 ounces

If 3 ounces spilled out the remaining amount of shampoo = ( 30.5 - 3 ) = 27.5 ounces.

Every time he shampoos his dog, amount of shampoo required = 1\frac{1}{4} ounces = 1.25 ounce

If x is the number of times Nico can bathe his dog using the same bottle of shampoo then total amount used = (1.25) (x) ounces

Now remaining amount of shampoo = ttal amount used to shampoo the dog

(27.5) = (1.25x)

x= \frac{27.5}{1.25}  = 22 times

So answer will be equation ( 1.25x = 27.5 ) and x = 22.

Ksju [112]3 years ago
4 0

Answer:

30 1/2 - 3 = 27 1/2

27 1/2 = 1 1/4x

he can bathe his dog 22 times

Step-by-step explanation:

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You work at a pioneer historical site. On this site you have handcarts. One cart has a handle that connects to the center of the
Gelneren [198K]

Answer:

a)  see below

b)  radius = 16.4 in (1 d.p.)

c)  18°. Yes contents will remain. No, handle will not rest on the ground.

d)  Yes contents would spill.  Max height of handle = 32.8 in (1 d.p.)

Step-by-step explanation:

<u>Part a</u>

A chord is a <u>line segment</u> with endpoints on the <u>circumference</u> of the circle.  

The diameter is a <u>chord</u> that passes through the center of a circle.

Therefore, the spokes passing through the center of the wheel are congruent chords.

The spokes on the wheel represent the radii of the circle.  Spokes on a wheel are usually evenly spaced, therefore the congruent central angles are the angles formed when two spokes meet at the center of the wheel.

<u>Part b</u>

The <u>tangent</u> of a circle is always <u>perpendicular</u> to the <u>radius</u>.

The tangent to the wheel touches the wheel at point B on the diagram.  The radius is at a right angle to this tangent.  Therefore, we can model this as a right triangle and use the <u>tan trigonometric ratio</u> to calculate the radius of the wheel (see attached diagram 1).

\sf \tan(\theta)=\dfrac{O}{A}

where:

  • \theta is the angle
  • O is the side opposite the angle
  • A is the side adjacent the angle

Given:

  • \theta = 20°
  • O = radius (r)
  • A = 45 in

Substituting the given values into the tan trig ratio:

\implies \sf \tan(20^{\circ})=\dfrac{r}{45}

\implies \sf r=45\tan(20^{\circ})

\implies \sf r=16.37866054...

Therefore, the radius is 16.4 in (1 d.p.).

<u>Part c</u>

The measure of an angle formed by a secant and a tangent from a point outside the circle is <u>half the difference</u> of the measures of the <u>intercepted arcs</u>.

If the measure of the arc AB was changed to 72°, then the other intercepted arc would be 180° - 72° = 108° (since AC is the diameter).

\implies \sf new\: angle=\dfrac{108^{\circ}-72^{\circ}}{2}=18^{\circ}

As the handle of the cart needs to be no more than 20° with the ground for the contents not to spill out, the contents will remain in the handcart at an angle of 18°.

The handle will not rest of the ground (see attached diagram 2).

<u>Part d</u>

This can be modeled as a right triangle (see diagram 3), with:

  • height = (48 - r) in
  • hypotenuse ≈ 48 in

Use the sin trig ratio to find the angle the handle makes with the horizontal:

\implies \sf \sin (\theta)=\dfrac{O}{H}

\implies \sf \sin (\theta)=\dfrac{48-r}{48}

\implies \sf \sin (\theta)=\dfrac{48-45\tan(20^{\circ})}{48}

\implies \theta = 41.2^{\circ}\:\sf(1\:d.p.)

As 41.2° > 20° the contents will spill out the back.

To find the <u>maximum height</u> of the handle from the ground before the contents start spilling out, find the <u>height from center of the wheel</u> (setting the angle to its maximum of 20°):

\implies \sin(20^{\circ})=\dfrac{h}{48}

\implies h=48\sin(20^{\circ})

Then add it to the radius:

\implies \sf max\:height=48\sin(20^{\circ})+45\tan(20^{\circ})=32.8\:in\:(1\:d.p.)

(see diagram 4)

------------------------------------------------------------------------------------------

<u>Circle Theorem vocabulary</u>

<u>Secant</u>: a straight line that intersects a circle at two points.

<u>Arc</u>: the curve between two points on the circumference of a circle

<u>Intercepted arc</u>: the curve between the two points where two chords or line segments (that meet at one point on the other side of the circle) intercept the circumference of a circle.

<u>Tangent</u>: a straight line that touches a circle at only one point.

7 0
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