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PLEASE help its for an exam i would really appreciate it , at the top it says “Drag the tiles to solve the given equation for x and justify step in your solution. you must fill every slot in the table in order for your answer to be scored correctly.”

Answers

Answer 1

Answer:

Answer:

2A. 2x-7 =3

B. Subtraction property

3A. 2x = 10

B. Addition property

4A. x=5

B. Division property

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Please help, i will give brainliest

Answers

Answer:

Step-by-step explanation:

hope i helped

Find atleast 5 numbers between 1/2 and 1/3

Answers

0.4 0.345 0.429 0.377 0.448

Hector's school is holding a fitness challenge. Student are encouraged to exercise at least 2 1/2 hours per week. Hector exercises about the same number of hours each week. During a 4-week period, he exercises for 11 1/2 hours. Hector wants to compare his exercise rate with the fitness challenge rate. How many hours per week does Hector exercise?

Answers

Answer:

The answer is 2 7/8

Final answer:

Hector exercises 2.875 hours per week, which is calculated by dividing the total of 11 1/2 hours he exercised during the 4 weeks by 4.

Explanation:

Hector wants to compare his exercise rate with the fitness challenge rate. To find out how many hours per week Hector exercises, we need to divide the total hours he exercised in a 4-week period by the number of weeks.

Hector exercised for 11 1/2 hours during 4 weeks, which is equal to 11.5 hours. Dividing 11.5 hours by 4 weeks gives us:

11.5 hours ÷ 4 weeks = 2.875 hours per week

Thus, Hector exercises 2.875 hours per week, which is slightly more than the 2 1/2 hours (or 2.5 hours) per week recommended by the fitness challenge.

Learn more about Exercise Rate here:

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The solution to 8x = 32 is x = __.

Answers

$\huge\text{Hey there!}$ !

$\large\text{8x = 32}$

$\large\text{DIVIDE 8 to BOTH SIDES}$

$\mathsf{(8x)/(8) = (32)/(8)}$

$\large\text{Cancel out: } \mathsf{(8)/(8)}\large\text{ because that gives you } \mathsf{1}$

$\large\text{Keep: }\mathsf{(32)/(8)}\large\text{ because it helps us solve for }\mathsf{x}$

$\mathsf{(32)/(2)\ = x \ or\ x = (32)/(8)}$

$\mathsf{(32)/(8) = 32/ 8 \rightarrow \bf 4}$

$\boxed{\boxed{\large\text{Answer: \bf x = 4}}}\huge\checkmark$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

A fine fabric was selling for $ 10.50 a yard. Joni bought 2 1/3 yards. How much did she spend?

Answers

Answer:

$24.50

Step-by-step explanation:

10.50 * 2 1/3 = 24.50

I would appreciate brainliest, have an amazing day

A trough has a semicircular cross section with a radius of 9 feet. Water starts flowing into the trough in such a way that the depth of the water is increasing at a rate of 2 inches per hour. (a) Give a function w = f(t) relating the width w of the surface of the water to the time t, in hours. Make sure to specify the domain and compute the range too.(b) After how many hours will the surface of the water have width of 6 feet?

(c) Give a function t = f −^1 (w) relating the time to the width of the surface of the water. Make sure to specify the domain and compute the range too.

Answers

Answer:

(a) Let h represents the height of water and w represents the width of the water,

Since, the depth of the water is increasing at a rate of 2 inches per hour,

So, after t hours,

The height of water, h(t) = 2t inches = t/6 ft,

( ∵ 1 foot = 12 inches ⇒ 1 inch = 1/12 ft )

Thus, the distance distance from the centre to the top of the water, d = 9 - h(t) ( see in the diagram )

$d=9-(t)/(6)$ ,

By the Pythagoras theorem,

$d^2 + ((w)/(2))^2 = 9^2$

$(9-(t)/(6))^2 +(w^2)/(4) = 81$

$(t^2)/(36)-(18t)/(6) + (w^2)/(4)=0$

$(t^2 - 108t + 9w^2)/(36)=0$

$t^2 - 108t + 9w^2 =0$

$9w^2 = 108t - t^2$

$w = (1)/(3)√(108t - t^2)$

Since, diameter of the semicircular cross section is 18 ft,

So, 0 ≤ w ≤ 18,

i.e Range = [0, 18]

Also, w will be defined if 108t - t² ≥ 0

⇒ (108 - t)t ≥ 0,

⇒ 0 ≤ t ≤ 108

i.e Domain = [0, 108]

(b) If w = 6,

$6 =(1)/(3)√(108t - t^2)$

$18 =√(108t-t^2)$

$324 = 108t - t^2$

$\implies t^2 - 108t+ 324=0$

By using quadratic formula,

$\implies t = 3.088\text{ or }t = 104.912$

Hence, After 3.1 hours or 104.9 hours will the surface of the water have width of 6 feet.

(c) $w = (1)/(3)√(108t- t^2)$

$\implies 3w = √(108t- t^2)$

$9w^2 = 108t - t^2$

$-9w^2 = -108t + t^2$

$-9w^2 + 2916 = 2916 - 108t + t^2$

$2916 - 9w^2 = (t - 108)^2$

$(t-108) = √(2916 - 9w^2)$

$t = √(2916 - 9w^2) + 108$

For 0 ≤ w ≤ 18,

0 ≤ t ≤ 108,

So, Domain = [0, 18]

Range = [0, 108]

Final answer:

The width of the water's surface in a semicircular trough can be represented by the function w=t/3 and its domain is t ≥ 0 and the range is 0 ≤ w ≤ 6. To have a 6 feet wide surface, thus, it would take 18 hours. The inverse function is t=3w, with a domain of 0 ≤ w ≤ 6 and range of t ≥ 0.

Explanation:

Given that the depth of the water is increasing at a rate of 2 inches per hour in a semicircular trough, we can convert this rate to feet per hour by dividing by 12, getting an increase of 1/6 feet per hour.

(a) We can express the width of the surface of the water as a function of time. We consider the cross-section of the trough is a semicircle. So, the radius of the water's surface will be the height of water, and this height increases at 1/6 feet per hour. Therefore, the width of the surface of water, w=2r=2*1/6t=t/3. The domain of the function is t ≥ 0 and the range is 0 ≤ w ≤ 6.

(b) We set w=6 in the function w=t/3 and solve for t. We get t=3*6=18 hours.

(c) The inverse function of w=t/3 is t=3w. The domain of the inverse function is 0 ≤ w ≤ 6 and the range is t ≥ 0.

Learn more about Mathematical functions here:

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