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pickupchik [31]
2 years ago
8

The figure below shows the ideal pattern of movement of a herd of cattle, with the arrows showing the movement of the handler as

he moves the herd. The arc the handler makes from the starting point to the return point should be a quarter of a circle:
A sector showing a quarter of a circle is drawn. The radius is marked as 70 feet. The endpoints of the arc of the sector are ma

Based on this theory, what distance will the handler move from the starting point to the return point if he creates an arc of a circle of radius 70 feet? (6 points)

439.6 feet
3846.5 feet
109.9 feet
1758.4 feet

Mathematics
2 answers:
yKpoI14uk [10]2 years ago
8 0

Answer:

c

Step-by-step explanation:

densk [106]2 years ago
8 0

Answer:

109.9 feet is correct

Step-by-step explanation:

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An explosion causes debris to rise vertically with an initial speed of 120 feet per second. The formula h equals negative 16 t s
Novay_Z [31]

Answer:

The debris will be at a height of 56 ft when time is <u>0.5 s and 7 s.</u>

Step-by-step explanation:

Given:

Initial speed of debris is, s=120\ ft/s

The height 'h' of the debris above the ground is given as:

h(t)=-16t^2+120t

As per question, h(t)=56\ ft. Therefore,

56=-16t^2+120t

Rewriting the above equation into a standard quadratic equation and solving for 't', we get:

-16t^2+120t-56=0\\\textrm{Dividing by -8 throughout, we get}\\\frac{-16}{-8}t^2+\frac{120}{-8}t-\frac{56}{-8}=0\\2t^2-15t+7=0

Using quadratic formula to solve for 't', we get:

t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(7)}}{2(2)}\\\\t=\frac{15\pm \sqrt{225-56}}{4}\\\\t=\frac{15\pm\sqrt{169}}{4}\\\\t=\frac{15\pm 13}{4}\\\\t=\frac{15-13}{4}\ or\ t=\frac{15+13}{4}\\\\t=\frac{2}{4}\ or\ t=\frac{28}{4}\\\\t=0.5\ s\ or\ t=7\ s

Therefore, the debris will reach a height of 56 ft twice.

When time t=0.5\ s during the upward journey, the debris is at height of 56 ft.

Again after reaching maximum height, the debris falls back and at t=7\ s, the height is 56 ft.

5 0
3 years ago
4 + 6^2 + 9 ∙ 3 - 10
Korvikt [17]
4+6^2+9•3-10

PEMDAS
P=parentheses
E=exponent
m=multiplication
d=division
a=addition
s=subtraction

*no parentheses in this equation
*exponent = 6^2
6^2=36
4+36+9•3-10
*multiplication = 9•3
9•3=27
4+36+27-10
*addition=4+36+27
4+36+27=67
67-10
*subtraction=67-10
67=10=57

ANSWER=57







please give brainliest itd mean a lot<3
3 0
3 years ago
Simplify the expression. ​
deff fn [24]

(6^-2)^2

When you have a number raised to a power inside parenthesis and then it is raised to another power, to simplify, multiply the two powers together:

-2 x 2 = -4

Simplified is 6^-4

5 0
3 years ago
Read 2 more answers
Given: ∆PQR, m∠R = 90° m∠PQR = 75° M ∈ PR , MP = 18 m∠MQR = 60° Find: RQ
tensa zangetsu [6.8K]

Answer:    RQ= 8.99 ( approx)

Step-by-step explanation:

Let MR= x

Since, In triangle, PRQ, tan 75°= \frac{18+x}{RQ}

⇒ RQ=  \frac{18+x}{tan 75^{\circ}}

Now, In triangle MRQ,

tan 60°= \frac{18+x}{RQ}

⇒ RQ=  \frac{x}{tan 60^{\circ}}

On equating both values of RQ,

\frac{18+x}{tan 75^{\circ}}=\frac{x}{tan 60^{\circ}}

⇒\frac{18+x}{x}=\frac{tan 75^{\circ}}{tan 60^{\circ}}

⇒\frac{18+x}{x}=\frac{tan 75^{\circ}}{tan 60^{\circ}}

⇒\frac{18+x}{x}=2.15470053838

⇒18=2.15470053838x-x

⇒x=15.5884572681≈15.60

Thus RQ=8.99999999999≈8.99


6 0
3 years ago
Select the graph that represents two quantities in a proportional relationship.
julia-pushkina [17]

Answer:

B

Step-by-step explanation:

6 0
3 years ago
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