An object is dropped (g = 9.80 m/s ^ 2) and falls for 5.00 s before hitting the ground. How fan has the object fallen

Answer:

its the 2nd one

hope that helps<3

Answer:

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Step-by-step explanation:

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Answer:

1. option c

2. option b

3. option b

Step-by-step explanation:

1 . option C

Square of longer side = sum of squares of other side

a)

5, 6 , 11

[tex]121 = 25 + 36\\does\ not \ satisfy[/tex]

b)

9, 12, 16

[tex]256 = 144 + 81\\Does\ not \ satisfy[/tex]

c)

7, 24, 25

[tex]625 = 576 + 49\\Does\ satisfy[/tex]

d)

12, 20, 25

[tex]625 = 400 + 144\\does \ not \ satisfy[/tex]

2. Option B

Square of longer side = sum of squares of other side

15 ² = 8 ² + r²

225 = 64 + r²

161 = r²

r = 12.688

r = 13ft

3 . Option b

To find length  of AB,  Use distance formula

[tex]AB = \sqrt{(x_2- x_1)^2 + (y_2 - y_1)^2}[/tex]

    [tex]= \sqrt{(14-3)^2 + (6-1)^2}\\\\=\sqrt{11^2 + 5^2}\\\\=\sqrt{121 + 25}\\\\= \sqrt{146} \\= 12 \ ft[/tex]

Answer:

[tex]sin 2 \theta = \frac{4 \sqrt5}{9}[/tex]

Step-by-step explanation:

Identities used :

[tex]1. sin^2 \theta = 1 - cos^2\theta\\\\2. tan^2 \theta + 1 = sec^2 \theta\\\\3.cos \theta = \frac{1}{sec \theta} \\\\4. sin 2 \theta = 2 sin\theta cos\theta[/tex]

[tex]Given \ tan \theta = -\frac{2 \sqrt5}{5}\\[/tex]

[tex]we \ know \ tan ^2 \theta + 1 = sec^2\theta[/tex]

            [tex](-\frac{2 \sqrt 5}{5})^2 + 1 = sec^2 \theta\\\\\frac{4 \times 5}{25 } + 1 = sec^2 \theta \\\\\frac{20 + 25}{ 25} = sec^2 \theta \\\\\frac{45}{25} = sec^2 \theta \\\\sec \theta = \sqrt{ \frac{45}{25}}\\\\sec \theta = \sqrt{ \frac{9 \times 5}{5 \times 5}} \\\\sec \theta = \frac{3 \sqrt {5}}{5}\\\\[/tex]

[tex]We \ know \ cos \theta = \frac{1}{ sec \theta}[/tex]

[tex]So , \cos \theta = \frac{5}{3 \sqrt5}[/tex]

[tex]Next \ sin^2 \theta = 1 - cos^2 \theta[/tex]

                 [tex]= 1 - (\frac{5}{3 \sqrt5})^2\\\\=1 - \frac{25}{9 \times 5}\\\\=1 - \frac{25}{45}\\\\=\frac{45 - 25}{45}\\\\=\frac{20}{45}\\\\=\frac{4}{9}[/tex]

    [tex]sin \theta = \sqrt{ \frac{4}{9}} = \frac{2}{3}[/tex]

[tex]Find \ sin 2 \theta[/tex]

[tex]sin 2 \theta = 2 (sin \theta \times cos \theta)[/tex]

        [tex]= 2 \times \frac{2}{3} \times \frac{5}{3\sqrt5}\\\\= \frac{4 \times \sqrt5 \times \sqrt5}{9 \times \sqrt 5}\\\\=\frac{4 \sqrt5}{9}[/tex]

9514 1404 393

Answer:

  (2ac+1)/(abc+2c+1)

Step-by-step explanation:

It helps to understand the factoring of the numbers involved.

  63 = 3²·7

  140 = 2²·5·7

The "change of base" formula is also useful.

  [tex]\log_b{a}=\dfrac{\log{a}}{\log{b}}[/tex]

__

Using this, we can write ...

  a = log(3)/log(2)   ⇒   log(3) = a·log(2)

  b = log(5)/log(3)   ⇒   log(5) = b·log(3) = ab·log(2)

  c = log(2)/log(7)   ⇒   log(2) = c·log(7)

This lets us write everything in terms of log(7):

  log(2) = c·log(7)

  log(3) = ac·log(7)

  log(5) = abc·log(7)

__

The desired logarithm is ...

  [tex]\log_{140}{}63=\dfrac{\log{63}}{\log{140}}=\dfrac{2\log{(3)}+\log{(7)}}{2\log{(2)}+\log{(5)}+\log{(7)}}\\\\=\dfrac{2ac\cdot\log{(7)}+\log{(7)}}{2c\cdot\log{(7)}+abc\cdot\log{(7)}+\log{(7)}}\\\\\boxed{\log_{140}{63}=\dfrac{2ac+1}{abc+2c+1}}[/tex]

_____

The attachment is a numerical check of this result.

Answer:

35

Step-by-step explanation:

To find this, we are going to use the combination formula as follows;

nCr = n!/(n-r)!r!

In this case, n = 7 and r = 3

Substituting these values;

7C3 = 7!/(7-3)! 3! = 7!/4!3! = 35

Step-by-step explanation:

(-) + (-) = (+)

=........ +1

Answer:

-1

Step-by-step explanation:

Answer:

Its B trustStep-by-step explanation

Step-by-step explanation:

Answer:

2⁴ × 3³ × 5

Step-by-step explanation:

By prime factorisation,

→ 2160 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5

→ 2160 = 2⁴ × 3³ × 5

Answer:

A. 20

B. 11 + 3tt

Step-by-step explanation:

A.  11 + 3(3) = 20

B.  11 + 3(tt)

Answer:

Sorry for the late response, I had to have lunch.

1. >

2. =

3. <

4. >

Step-by-step explanation:

Have a great summer :)

Answer:

T = 3.5m

T is trash in pounds

m is members of family

Step-by-step explanation:

Answer:

x = 25

Step-by-step explanation:

Correct question

In a right triangle Sin (40-x)=cos(3x) what is the value of x

Since Sin theta = Cos(90-theta)

Sin (40-x) = Sin(90-3x)

40 - x = 90 - 3x

-x+3x = 90 - 40

2x = 50

x = 25

Hence the value of x is 25

a) Two angles are said to be complementary if their sum is equal to 90°

Given : 2x and 4x are complementary angles

⇒   2x + 4x = 90°

⇒   6x = 90°

⇒   x = 15°

Answer: The two angles are : 30° and 60°

b) Two angles are said to be supplementary if their sum is equal to 180°

Given : x and 5x are supplementary angles

⇒   x + 5x = 180°

⇒   6x = 180°

⇒   x = 30°

Answer: The two angles are : 30° and 150°

c) Two angles are said to be complementary if their sum is equal to 90°

Given : x and x + 50° are complementary angles

⇒   x + x + 50° = 90°

⇒   2x = 40°

⇒   x = 20°

Answer: The two angles are : 20° and 70°

d) If two angles form a linear pair, then the sum of the angles should be equal to 180°

Given : 2x + 10° and 3x + 10° form a linear pair

⇒   2x + 10° + 3x + 10° = 180°

⇒   5x = 160°

⇒   x = 32°

⇒  2x + 10° = 2(32) + 10° = 74°

⇒  3x + 10° = 3(32) + 10° = 106°

Answer: The two angles are 74° and 106°

Answer:

2300 mm^2

Step-by-step explanation:

Answer:

B and C

Step-by-step explanation:

B is correct because 5² is 25, which is less than 64

C is correct because (-3)² is 9, which is less than 64

Answer:

b c d f

Step-by-step explanation:

-8 ≤ x ≤ 8

Answer:

23,300 students

Step-by-step explanation:

1900 year, 3.7 * 10 ^ 3

3.7 * 10 * 10 * 10 = 3,700

1950 year,  2.7 * 10 ^ 4

2.7 * 10 * 10 * 10 * 10 = 27,000

So then you will subtract 1950 year by the year of 1900

27,000 - 3,700 = 23,300

Answer:

8

Step-by-step explanation:

a + b^3 - a

Simplify

b^3

Let b=2

2^3

8

Answer:

11

Step-by-step explanation:

If a = 3, and b = 2, then;

=> a + b³

=> 3 + 2³

=> 3 + 8

=> 11

Hope this helps!

Answer:

13

Step-by-step explanation:

The IQR = Interquartile Range is the difference between the first and third quartile

Rearranged data set:

3, 7, 16, 17, 20, 45

Step 1

We find the first quartile = Q1

= 1/4(n + 1)th number

n = 6

= 1/4(6 + 1)th

= 1/4(7)th

= 1 3/4th number

This = 7

Step 2

We find the third quartile = Q3

= 3/4(n + 1)th number

n = 6

= 3/4(6 + 1)th

= 3/4(7)th

= 21/4th number

= 5 1/4 th number

This = 20

Step 3

IQR = Q3 - Q1

= 20 - 7

= 13

Answer:

39.28cm^2

Step-by-step explanation:

Simple work out the area of a circle and halve it.

3.142xr^2

3.142x5^2 = 78.55 cm^2 (Full circle area)

Now we must halve that answer to get the semicircle.

78.55/2 = 39.275 cm^2

However, we need it to 2dp. Therefore the answer is 39.28cm^2

Answer:

16 boxes large

Step-by-step explanation:

if the weigth is 1280 wilhich means its 1280÷80 =16

Answer:

Perimeter: 8x+8

Area: 16x+16

Step-by-step explanation:

Combine only like terms

(4x+2) + (4x+2) +2+2= 8x+8

(4x+2) x (4x+2) x 2 x 2= 16x+16

Answer:

Step-by-step explanation:

A?

Answer:

35 is correct

Step-by-step explanation:

hope this helps.

Answer:

35°

Step-by-step explanation:

Angle ACD is 70°. The little tick marks on the angle mean both sides of the split are the same amount. This means if you divide the angle in half, you will find out what both of them are equal to.

Answer:

Step-by-step explanation:

[tex] {10}^{ - 3} [/tex]

Answer:

Compressions

Step-by-step explanation:

The third type is compressions

Answer:

288 cubic feet

Step-by-step explanation:

add the 3 blocks.

block 1: 12*6*2=12*12=144

block 2: (12-4)*2*6=8*2*6=8*12=96

block 3: (12-4-4)*6*2=4*12=48

add them up:

144+96+48

288 cubic feet

with help of calculator

Answers:

He has  27   quarters and  31   dimes

=========================================================

Work Shown:

d = number of dimes

q = number of quarters

d+q = 58 since there are 58 coins of only these two types

d = 58-q

10d+25q = 985 .... note that 985 cents is $9.85

10(58-q)+25q = 985 .... replace d with 58-q

580-10q+25q = 985

15q+580 = 985

15q = 985-580

15q = 405

q = 405/15

q = 27 ..... Mike has 27 quarters

d = 58-q

d = 58-27

d = 31 .... Mike also has 31 dimes

-----------

As a way to check:

d+q = 31+27 = 58 so that works out

and

10d+25q = 10*31+25*27 = 310+675 = 985

so that works out as well. The answer is confirmed.

Answer:

In picture

Step-by-step explanation:

Hope I answered it correctly.

Please give me brainliest if it help you, thanks!

Answer:

Triangle A:

- smaller leg = 3

- longer leg = 10

- hypotenuse = 10.4

Traingle B:

- smaller leg = 4

- longer leg = 13

- hypotenuse = 13.6

Triangle C:

- smaller leg = 5

- longer leg = 7

- hypotenuse = 8.6

Step-by-step explanation:

Let's find the distance of the legs for the 3 right triangles.

Triangle A

The smaller leg (SL) is from (2,5) to (2,8), so the length is given by the distance in the y-axis:

[tex] SL_{A} = 8 - 5 = 3 [/tex]

Hence, the lenght of the smaller leg is 3.

The longer leg (LL) is from (-8, 8) to (2, 8), so the length is given by the distance in the x-axis:

[tex] LL_{A} = 2 - (-8) = 10 [/tex]

Then, the length of the longer leg is 10.  

Now, we can find the hypotenuse (H) by using Pitagoras:

[tex] H_{A} = \sqrt{SL_{A}^{2} + LL_{A}^{2}} = \sqrt{3^{2} + 10^{2}} = 10.4 [/tex]

Triangle B

The SL is from (3, -9) to (7, -9)

The length of the SL is:

[tex] SL_{B} = 7 - 3 = 4 [/tex]

The LL is from (7, -9) to (7, 4)

The length of the LL is:

[tex] LL_{B} = 4 - (-9) = 13 [/tex]

The hypotenuse is:

[tex] H_{B} = \sqrt{4^{2} + 13^{2}} = 13.6 [/tex]

Triangle C

The SL is from (-10, -6) to (-10, -1)

The length of the SL is:

[tex] SL_{C} = -1 - (-6) = 5 [/tex]

The LL is from (-10, -6) to (-3, -6)  

The length of the LL is:

[tex] LL_{C} = -3 - (-10) = 7 [/tex]

The hypotenuse is:

[tex] H_{C} = \sqrt{5^{2} + 7^{2}} = 8.6 [/tex]

I hope it helps you!

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