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2 years ago

12

I need help what is y?

1 answer:

Darina [25.2K]2 years ago

3 0

Answer:

$y=-5.6$

Step-by-step explanation:

$2.5\left(y + \dfrac{2}{5} \right) =-13\\\\\textsf{divide both sides by 2.5} : y+ \dfrac{2}{5}=-5.2\\\\\textsf{subtract} \ \dfrac{2}{5} \ \textsf{from both sides} : y=-5.6$

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On Monday Ravi drives for 4 hours.

tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

On Monday Ravi drives for 4 hours.

His average speed is 30 mph.

How far does Ravi drive on Monday?

4 0

3 years ago

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PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!!

Elenna [48]

Answer:

$cos(V)=\frac{39}{89}=.44$

∠ $V=64.01$ °

Step-by-step explanation:

$sin(angle)=\frac{opposite}{hypotenuse}$

$cos(angle)=\frac{adjacent}{hypotenuse}$

$tan(angle)=\frac{opposite}{adjacent}$

The cosine of an angle is equal to the side adjacent to the angle over the hypotenuse, the longest side.

With this in mind:

$cos(V)=\frac{39}{hypotenuse}$

We need to find the hypotenuse of this triangle. This is relativly easy to do with a right triangle; we can just use the Pythagorean Theorem.
Pythagorean Theorem: $a^2+b^2=c^2$ where $a$ , $b$ , and $c$ are the 3 different sides of a right triangle.

$a^2+b^2=c^2\\(39)^2+(80)^2=c^2\\1521+6400=c^2\\c=\sqrt{7921} \\c=89$

Our hypotenuse is $89$ . Now we can finish solving for ∠ $V$ .

$cos(V)=\frac{39}{89}\\V=cos^-^1(\frac{39}{89})\\$

∠ $V=64.01$ °

3 0

4 years ago

At the local sandwich shop, employees earn double time for any hours worked in addition to their regular 40-hour week. Last week

Ede4ka [16]

Double-time for overtime...so overtime pay is 2(8.50) = 17

40(8.50) + 9(17) = 340 + 153 = 493 <== pay for the week

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4 years ago

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A force of 7 lb is required to hold a spring stretched 9 in. beyond its natural length. How much work W is done in stretching it

Ymorist [56]

144 joules is the work W done in stretching it from its natural length to 13 in beyond its natural length

Force F=10 lb

From the hookes Law

$F=kx$

Therefore, calculate k for the spring for

$10=k(\frac{4}{12}) = > k=\frac{120}{4}=30$

Work done in stretching a spring through a length dx is

$dW=F.dx = > dW=kxdx$

For calculating the work done for stretching the spring to x=6in=(6/12)feet beyond the natural length, Integrate over the limits of x=0 to x=1/2

Therefore,

$\int_{0}^{W}dW=\int_{0}^{0.5}kxdx = > W=\frac{1}{2}k[x^2]_{0}^{0.5}=\frac{1}{2}30 \times (0.5)^2=\frac{30}{8}$

That is,

$W=\frac{15}{4} ft-lb\\$

The answer that is given is for stretching the spring to 4 inches.

Mass of 10 m of chain length is 80kg. This implies, Mass per unit meter length of the chain is

$m_{l}=\frac{80}{10}=8kg/m$

Consider a small length dx of the chain at the end point A. Work done in lifting the small length dx over the height of x meters is

$dW=(8g)dx \times x=8gxdx$

Now, integrate over the the value of x from

x=0 when end of the chain A is on the ground

x=6 when end of the chain A reaches 6 m above the ground

That is,

$\int_{0}^{W}dW=\int_{0}^{6}8gxdx=8g [\frac{x^2}{2}]_{0}^{6}=4g(6^2-0^2)$

$W=4g \times 36=144g \ Joules$

Hence,144 joules is the work W done in stretching it from its natural length to 13 in beyond its natural length

Learn more about Integration here brainly.com/question/2263647

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7 0

2 years ago

The average wind speed at the weather station of a certain city is 34.4 miles per hour. The

klio [65]

196.9 miles

231.3 miles - 34.4 miles = 196.9 miles

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4 years ago

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