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3 years ago

A 150-N bird feeder is supported by three cables. Find the tension in each cable.

Mathematics

2 answers:

Irina-Kira [14]3 years ago

6 0

X axis : -t1 cos 60 + t2 cos 60 = 0

t2 cos 60 = t1 cos 60 ; t2 = t1

Y axis : t1 sin 60 + t2 sin 60 - 150 = 0

since t2 = t1

2t1 sin 60 = 150

t1 ( sqrt(3)/2) = 150/2

t1 = 50 [sqrt(3)]

t2 = 50 [sqrt(3)]

t3 = 150 N

hope this helps

Klio2033 [76]3 years ago

5 0

see the attached figure with letters to better understand the problem

we know that

in the x-coordinates

$TACx=TBCx\\ TAC*cos60=TBC*cos30\\\\ TAC*\frac{1}{2} =TBC*\frac{\sqrt{3}}{2}$

$TAC=TBC*\sqrt{3}$ -----> equation $1$

in the y-coordinates

$TCD=150 N$

$TACy+TBCy=TCD\\ TAC*sin60+TBC*sin 30=150\\ \\ TAC*\frac{\sqrt{3}}{2} +TBC*\frac{1}{2} =150\\ \\ TAC*\sqrt{3} +TBC=300$

$TAC*\sqrt{3} +TBC=300$ -----> equation $2$

substitute equation $1$ in equation $2$

$[TBC*\sqrt{3} ]*\sqrt{3} +TBC=300\\ 3TBC+TBC=300\\\\ TBC= \frac{300}{4} \\ \\ TBC=75 N$

$TAC=TBC\sqrt{3}\\ TAC=75\sqrt{3} N$

therefore

the answer is

$TAC=75\sqrt{3} N\\ TBC=75N\\ TCD=150N$

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saw5 [17]

Answer:

The junior class will receive US$ 1,479.77 when it cashes in the Certificate of Deposit.

Step-by-step explanation:

1. Let's review the data given to us for answering the question:

Amount invested in the Certificate of Deposit = US$ 1,400

Duration of the CD = 18 months

Interest rate = 3.7% compounded monthly

2. Let's find the future value of the Certificate of Deposit after 18 months, using the following formula:

FV = PV * (1 + r) ⁿ

PV = Amount invested = US$ 1,400

number of periods (n) = 18

rate (r) = 3.7% = 0.037 annually or 0.003083 monthly

Replacing with the real values, we have:

FV = 1,400 * (1 + 0.003083) ¹⁸

FV = 1,400 * (1.003083) ¹⁸

FV = 1,400 * 1.056978

FV = US$ 1,479.77 (Rounding to the nearest cent)

The junior class will receive US$ 1,479.77 when it cashes in the Certificate of Deposit.

8 0

3 years ago

PLEASE HELPPP!!!

Tems11 [23]

Answer:

a) $-16t^{2}+64t\geq40$

b) $[2-\frac{\sqrt{6}}{2} , 2+\frac{\sqrt{6}}{2}]$

(see attached picture for the number line)

This interval means that from about 0.78s to about 3.22s the ball will not be visible because it will be higher than 40 feet high.

c) $h_{max}=64ft$

d) t=2s

Step-by-step explanation:

Basically the ball will disappear when it is higher than 40 ft, so we can build the inequality like this:

$-16t^{2}+64t\geq40$

this is because the function represents it's height, so when the function is greater than or equal to 40, then the ball will disappear.

b) to solve the inequality, first we need to turn the ≥ symbol to an = symbol and solve for t, so we get:

$-16t^{2}+64t =40$

we can move the 40 to the other side of the equation so we get:

$-16t^{2}+64t-40=0$

we can factor the equation so it's easier to solve, so we get:

$-8(2t^{2}-8t+5) =0$

which simplifies to:

$2t^{2}-8t+5=0$

and we can use the quadratic formula to solve this equation, so we get:

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

so we get:

$t=\frac{-(-8)\pm\sqrt{(-8)^{2}-4(2)(5)}}{2(2)}$

which yields:

$2\pm \frac{\sqrt{6}}{2}$

and next we need to test the posible intervals to see which one makes the inequality true, the possible intervals are:

$(-\infty, 2-\frac{\sqrt{6}}{2}]$ for a test value of 0

$[2-\frac{\sqrt{6}}{2},2+\frac{\sqrt{6}}{2}]$ for a test value of 1

$[2+\frac{\sqrt{6}}{2},\infty)$ for a test value of 4

so next we test each of the values in the original inequality and see if the inequality is true:

$(-\infty, 2-\frac{\sqrt{6}}{2}]$ for a test value of 0

$-16(0)^{2}+64(0)\geq40$

$0\geq40$ false

so this is not an answer.

$[2-\frac{\sqrt{6}}{2},2+\frac{\sqrt{6}}{2}]$ for a test value of 1

$-16(1)^{2}+64(1)\geq40$

$48\geq40$

true, so this is an answer.

$[2+\frac{\sqrt{6}}{2}, \infty)$ for a test value of 4

$-16(4)^{2}+64(4)\geq40$

$-16(16)+64(4)\geq40$

$-256+256\geq40$

$0\geq40$

false, so this is no answer.

So the answer is the interval:

$[2-\frac{\sqrt{6}}{2},2+\frac{\sqrt{6}}{2}]$

This interval means that from about 0.78s to about 3.22s the ball will not be visible because it will be higher than 40 feet high.

c) In order to find how high Penelope kick the football, we must find the vertex of the parabola which is gotten when graping the given function (see attached picture)

so we can use the following formula to find the t-value for this maximum height:

$t=-\frac{b}{2a}$

so we substitute the corresponding values:

$t=-\frac{64}{2(-16)}$

which yields:

t=2s

next, we can substitute this time into the original function to find the maximum height, so we get:

$h=-16(2)^{2}+64(2)$

which yields:

h=64 ft

d) We found the answer for d on the previous part where we got that t=2s

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3 years ago

Please Help me solve for x 3x-m=10

Triss [41]

Answer:

x=(10+m)/3

Step-by-step explanation:

3x-m=10

3x=10+m

x=(10+m)/3

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3 years ago

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The measure of angle A is 4 degrees greater than the measure of angle B. The two angles are complementary. Find the measure for

svetlana [45]

Given:
Angle A is 4 degrees greater than the measure of Angle B. Both angles are complementary

Complementary angles have a sum of 90°

Angle A = x + 4° ; Angle B = x

x + 4° + x = 90°
2x = 90° - 4°
2x = 86°
x = 86° ÷ 2
x = 43° ANGLE B.

Angle A = x + 4° ⇒ 43° + 4 = 47°

Given:
Angle D is 5 times the measure of Angle E. These angles are supplementary. This means that their sum is 180°

Angle D = 5x ; Angle E = x

5x + x = 180°
6x = 180°
x = 180° ÷ 6
x = 30° Angle E.

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fredd [130]

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If you move 2 units to the right (x-axis) you go from -6 + 2 which equals -4. And again, you’re not moving up and down so your
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