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MrRa [10]
2 years ago
9

Samir needs 3/8 yard of a string for a project. He has a stick that is 1/8 long. How many 1/8 yard lengths does he need to measu

re to get 3/8 yard of string
Mathematics
1 answer:
Katen [24]2 years ago
5 0

Answer:

1/8 adry

Step-by-step explanation:

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Draw a segment perpendicular to each given segment. Show your thinking by sketching triangles as needed.
Digiron [165]

Answer:

pls dont kill me for wacky lines. this shall just give you a basic idea. the important part is the 90° angle

7 0
3 years ago
PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU B
OleMash [197]

Answer:

X=3

Step-by-step explanation:

Hope this helps and have a wonderful day!!!

4 0
2 years ago
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
2 years ago
Question 8 Find the unit vector in the direction of (2,-3). Write your answer in component form. Do not approximate any numbers
slamgirl [31]

Answer:

The unit vector in component form is \hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}}  \right) or \hat{u} = \frac{2}{\sqrt{13}}\,i-\frac{3}{13}\,j.

Step-by-step explanation:

Let be \vec u = (2,-3), its unit vector is determined by following expression:

\hat {u} = \frac{\vec u}{\|\vec u \|}

Where \|\vec u \| is the norm of \vec u, which is found by Pythagorean Theorem:

\|\vec u\|=\sqrt{2^{2}+(-3)^{2}}

\|\vec u\| = \sqrt{13}

Then, the unit vector is:

\hat{u} = \frac{1}{\sqrt{13}} \cdot (2,-3)

\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}}  \right)

The unit vector in component form is \hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}}  \right) or \hat{u} = \frac{2}{\sqrt{13}}\,i-\frac{3}{13}\,j.

6 0
3 years ago
Write the point-slope equation oft line that goes through (1, -1) with slope of -5.​
Sever21 [200]

Answer:

y+1=-5(x-1)

Step-by-step explanation:

We need to find the slope intercept form first. So far we have this: y=-5x+b. We could plug in the x value and y value to get this : -1=-5(1)+b. b=4. Now we have the slope intecept form: y=-5x+4, and we could turn it into point slope form.

6 0
2 years ago
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