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3 years ago

To rent a certain meeting room, a college charges a reservation fee of

1 answer:

5 0

T = # of hours
Make inequality: 6.6t + 41 < 87.2
Subtraction: 6.6t < 46.2
Divide: t < 7
That means that the chemistry club can use the meeting room for less than 7 hours.

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Only need number 15 ugh i’m stupid

frutty [35]

Yes because 10^2+10^2=200
If you multiply a square root by itself it removes the root. Therefore making the statement 100+100=200; True

3 0

3 years ago

A point H is 20m away from the foot of a tower on the same horizontal ground. From the point H, the angle of elevation of the po

astra-53 [7]

Answer:

a. See Attachment 1

b. $PT = 12.3\ m$

c. $HT = 31.1\ m$

d. $OH = 28.4\ m$

Step-by-step explanation:

Calculating PT

To calculate PT, we need to get distance OT and OP

Calculating OT;

We have to consider angle 50, distance OH and distance OT

The relationship between these parameters is;

$tan50 = \frac{OT}{20}$

Multiply both sides by 20

$20 * tan50 = \frac{OT}{20} * 20$

$20 * tan50 = OT$

$20 * 1.1918 = OT$

$23.836 = OT$

$OT = 23.836$

Calculating OP;

We have to consider angle 30, distance OH and distance OP

The relationship between these parameters is;

$tan30 = \frac{OP}{20}$

Multiply both sides by 20

$20 * tan30 = \frac{OP}{20} * 20$

$20 * tan30 = OP$

$20 * 0.5774= OP$

$11.548 = OP$

$OP = 11.548$

$PT = OT - OP$

$PT = 23.836 - 11.548$

$PT = 12.288$

$PT = 12.3\ m$ (Approximated)

--------------------------------------------------------

Calculating the distance between H and the top of the tower

This is represented by HT

HT can be calculated using Pythagoras theorem

$HT^2 = OT^2 + OH^2$

Substitute 20 for OH and $OT = 23.836$

$HT^2 = 20^2 + 23.836^2$

$HT^2 = 400 + 568.154896$

$HT^2 = 968.154896$

Take Square Root of both sides

$HT = \sqrt{968.154896}$

$HT = 31.1\ m$ (Approximated)

--------------------------------------------------------

Calculating the position of H

This is represented by OH

See Attachment 2

We have to consider angle 50, distance OH and distance OT

The relationship between these parameters is;

$tan50 = \frac{OH}{OT}$

Multiply both sides by OT

$OT * tan50 = \frac{OH}{OT} * OT$

$OT * tan50 = {OH$

$OT * 1.1918 = OH$

Substitute $OT = 23.836$

$23.836 * 1.1918 = OH$

$28.4= OH$

$OH = 28.4\ m$ (Approximated)

5 0

2 years ago

If EFG ≅ △ ABD so FG is = .......? help asap

Igoryamba

Answer:

FG=BD

Step-by-step explanation:

hope this helps!!

5 0

2 years ago

Read 2 more answers

Given the Arithmetic series A1+A2+A3+A4 7 + 11 + 15 + 19 + . . . + 91 What is the value of sum?

aivan3 [116]

Answer:

The value of the sum is 1078

Step-by-step explanation:

* Lets revise how to find the sum of the arithmetic series

- In the arithmetic series there is a constant difference between each

two consecutive terms

- Ex:

# 4 , 9 , 14 , 19 , 24 , .......... (+ 5)

# 25 , 15 , 5 , -5 , -15 , .......... (-10)

- So if the first term is a and the constant difference between each two

consecutive terms is d, then

U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , ..........

- Then the nth term is Un = a + (n - 1)d

- The sum of n terms is Sn = n/2[a + L] , where L is the last term in

the series

* Lets solve the problem

∵ The arithmetic series is 7 + 11 + 15 + 19 + ......................... + 91

∴ The first term (a) is 7

∴ The last term is (L) 91

- Lets find the constant difference

∵ 11 - 7 = 4 and 15 - 11 = 4

∴ The constant difference (d) is 4

- Lets find the sum of the series from 7 to 91

∵ Sn = n/2[a + L]

∵ a = 7 and L = 91

∴ Sn = n/2[7 + 91]

- We need to know how many terms in the series

∵ L is the last term and equals 91 lets find its position in the series

∵ Un = a + (n - 1)d

∵ a = 7 , d = 4 and Un = 91

∴ 91 = 7 + (n - 1)(4) ⇒ subtract 7 from both sides

∴ 84 = (n - 1)(4) ⇒ divide both sides by 4

∴ 21 = n - 1 ⇒ add 1 to both sides

∴ n = 22

∴ The number of the terms in the series is 22

- Lets find the sum of the 22 terms (S22)

∴ S22 = 22/2[7 + 91]

∴S22 = 11[98] = 1078

* The value of the sum is 1078

4 0

3 years ago

Two boxes have the same volume. One box has a base that is 5\text{ cm}5 cm5, start text, space, c, m, end text by 5\text{ cm}5 c

sveticcg [70]

Answer:

4times tall

Step-by-step explanation:

Volume of the boxes = Base area × height

Volume of the first box V1 = A1h1

Given the base of the first box to be 5cm, the base area:

A1 = 5cm×5cm = 25cm²

Volume of the first box V1 = 25h1... 1

Similarly, volume of the second box

V2 = A2h2

Given the base of the second box to be 10cm, the base area:

A2= 10cm×10cm = 100cm²

Volume of the second box

V2 = 100h2... 2

If the two boxes have the same volume, then V1 = V2

25h1 = 100h2

divide both sides by 25

25h1/25 = 100h2/25

h1 = 4h2

Since the height of the smaller box is represented as h1, then the height of the smaller base is 4 times tall.

3 0

3 years ago