If a person high jumps 6 feet 2 inches, how many inches is the jump?

Answer:

108 square metres

Step-by-step explanation:

A=√d square - l square

here

A = area

d= diagonal

l= length

I think you are asked to find the value of the first derivative of f(x) at π/4. Given

[tex]f(x) = \dfrac{\sin(4x)-\cos(x)}{\tan(x)}[/tex]

use the quotient to differentiate and you get

[tex]f'(x) = \dfrac{\tan(x)(4\cos(4x)+\sin(x))-(\sin(4x)-\cos(x))\sec^2(x)}{\tan^2(x)}[/tex]

Then at x = π/4, you have

tan(π/4) = 1

cos(4•π/4) = cos(π) = -1

sin(π/4) = 1/√2

sin(4•π/4) = sin(π) = 0

cos(π/4) = 1/√2

sec(π/4) = √2

==>   f ' (π/4) = (1•(-4 + 1/√2) - (0 - 1/√2)•(√2)²) / 1² = -4 + 1/√2 + √2

Step-by-step explanation:

here is the answer. Feel free to ask for more.

Answer:

2 feet

Step-by-step explanation:

Displacement at 0 seconds is 0 feet.

Displacement at 2 seconds is 2 feet because it took them 2 seconds to walk 2 feet.

Answer:

283 flowers

Step-by-step explanation:

c=2pi*r

c = 1130.973 =1131

1131/4

282.75 = 283

Answer:

[tex]{ \tt{rate \: of \: change \: in \: A = 9}}[/tex]

Rate of change in function A is two times than that in function B

Answer:

x= - 1

Step-by-step explanation:

Answer:

Addition equation = -4-0) + [(-13)-(-4)]

Answer =  -13

Step-by-step explanation:

For the small arrow in the diagram, the expression is (-4 - 0)

For the bog arrow, the expression will be -13 - (-4)

Adding both expressions

Addition = (-4-0) + [(-13)-(-4)]

Addition = (-4) + (-13+4)

Addition = -4 + (-9)

Addition = -4-9

Addition = -13

Answer:

To find the surface area of a cuboid we can also label the length, width and the height of the prism and use the formula SA=2LW+2LH+2HW to find the area of a cuboid

Answer:

202 cm²

Step-by-step explanation:

The opposite faces of a cuboid are congruent , then

SA = top/bottom + front/ back + sides , that is

SA = 2(9 × 4) + 2(9 × 5) + 2(4 × 5)

     = 2(36) + 2(45) + 2(20)

     = 72 + 90 + 40

    = 202 cm²

Answer:

121 unit^2.

Step-by-step explanation:

The area = height/2 * ( sum of the opposite parallel  lines)

= h/2(BC + AD

h = BF = 14 - 3 = 11 units.

BC   = 13 - 5 = 8 units.

AD = 16 - 2 = 14 units.

Area = (11/2)(8 + 14)

= 5.5 * 22

= 121 unit^2.

Answer:

121

Step-by-step explanation:

Answer:

In any quantitative science, the terms relative change and relative difference are used to compare two quantities while taking into account the "sizes" of the things being compared. The comparison is expressed as a ratio and is a unitless number.

Step-by-step explanation:

I hope it helps

Answer:

22608 mm³/s

Step-by-step explanation:

Applying chain rule,

dV/dt = (dV/dr)(dr/dt)............... Equation 1

Where dV/dr = rate at which the volume is increasing

But,

V = 4πr³/3

Therefore,

dV/dr = 4πr²............... Equation 2

Substitute equation 2 into equation 1

dV/dt = 4πr²(dr/dt).............. Equation 3

From the question,

Given: dr/dt = 2 mm/s, r = 60/2 = 30 mm

Consatant: π = 3.14

Substitute these values into equation 3

dV/dt = 4×3.14×30²×2

dV/dt = 22608 mm³/s

Answer:

f(6)=-42

f(0)=0

f(-1)=7

Step-by-step explanation:

since f(x) =-7x

it implies that f(6),f(0) and f(-1) are all equal to f(x)

which is equal to -7x .

so wherever you find x you out it in the

function

Actual data table :

X __ y

2 15

6 13

7 9

8 8

12 5

Answer:

0.953

Step-by-step explanation:

The question isnt well formatted :

The actual data:

X __ y

2 15

6 13

7 9

8 8

12 5

Using a correlation Coefficient calculator, the correlation Coefficient obtained by fitting the data is 0.953 which depicts a strong linear correlation between the x and y variable. This shows that the value of y increases with a corresponding increase in x values and vice versa.

Answer: [tex]t\in [\dfrac{1}{4},2][/tex]

Step-by-step explanation:

Given

Inequality is [tex]4t^2\leq9t-2[/tex]

Taking variables one side

[tex]\Rightarrow 4t^2-9t+2\leq0\\\Rightarrow 4t^2-8t-t+2\leq0\\\Rightarrow 4t(t-2)-1(t-2)\leq0\\\Rightarrow (4t-1)(t-2)\leq0[/tex]

Using wavy curve method

[tex]t\in [\dfrac{1}{4},2][/tex]

Hi!

√80 = √(16 • 5) = √(4² • 5) = 4√5

(1) Recall that

sin(x - y) = sin(x) cos(y) - cos(x) sin(y)

sin²(x) + cos²(x) = 1

Given that α lies in the third quadrant, and β lies in the fourth quadrant, we expect to have

• sin(α) < 0 and cos(α) < 0

• sin(β) < 0 and cos(β) > 0

Solve for cos(α) and sin(β) :

cos(α) = -√(1 - sin²(α)) = -3/5

sin(β) = -√(1 - cos²(β)) = -5/13

Then

sin(α - β) = sin(α) cos(β) - cos(α) sin(β) =  (-4/5) (12/13) - (-3/5) (-5/13)

==>   sin(α - β) = -63/65

(2) In the second identity listed above, multiplying through both sides by 1/cos²(x) gives another identity,

sin²(x)/cos²(x) + cos²(x)/cos²(x) = 1/cos²(x)

==>   tan²(x) + 1 = sec²(x)

Rewrite the equation as

3 sec²(x) tan(x) = 4 tan(x)

3 (tan²(x) + 1) tan(x) = 4 tan(x)

3 tan³(x) + 3 tan(x) = 4 tan(x)

3 tan³(x) - tan(x) = 0

tan(x) (3 tan²(x) - 1) = 0

Solve for x :

tan(x) = 0   or   3 tan²(x) - 1 = 0

tan(x) = 0   or   tan²(x) = 1/3

tan(x) = 0   or   tan(x) = ±√(1/3)

x = arctan(0) + nπ   or   x = arctan(1/√3) + nπ   or   x = arctan(-1/√3) + nπ

x = nπ   or   x = π/6 + nπ   or   x = -π/6 + nπ

where n is any integer. In the interval [0, 2π), we get the solutions

x = 0, π/6, 5π/6, π, 7π/6, 11π/6

(3) You only need to rewrite the first term:

[tex]\dfrac{\sin(x)}{1-\cos(x)} \times \dfrac{1+\cos(x)}{1+\cos(x)} = \dfrac{\sin(x)(1+\cos(x))}{1-\cos^2(x)} = \dfrac{\sin(x)(1+\cos(x)}{\sin^2(x)} = \dfrac{1+\cos(x)}{\sin(x)}[/tex]

Then

[tex]\dfrac{\sin(x)}{1-\cos(x)}+\dfrac{1-\cos(x)}{\sin(x)} = \dfrac{1+\cos(x)+1-\cos(x)}{\sin(x)}=\dfrac2{\sin(x)}[/tex]

Answer:

D

Step-by-step explanation:

There 2 ways to interpret this problem.

From the info given:

These two triangles are congruent by SSS and congruent triangles have congruent or equal side lengths so the answer have to be 1.

If the triangles are similar, the side lengths form a proportion of that

[tex] \frac{3}{3} = \frac{3}{3} [/tex]

So the ratio or scale factor is 1.

The scale factor in the figure is 1.

A scale factor in math is the ratio between corresponding measurements of an object and a representation of that object.

Given that two triangles, LMN and OPQ, we need to find the scale factor,

We can see triangles are congruent, and we know that

Two triangles are congruent, by the SSS congruence criterion, if they are similar and the scale factor happens to be 1,

Hence, the scale factor in the figure is 1.

Learn more about scale factors, click;

https://brainly.com/question/29464385

#SPJ5

Answer:

A and b Are in quadrant 2. F and D are in quadrant 1. F is in quadrant 3 and C is in quadrant 4

Step-by-step explanation:

each quadrant is in the boxes and the question is asking what is each  coordinate is in what quadrant

Common difference: 6

First term: 7

Second term: 13

Third term: 19

Fourth term: 25

Fifth term: 31

I hope this is correct and helps!

Answer to the following question is as follows;

Number of term in AP (N) = 10

Step-by-step explanation:

Given:

Sum of arithmetic progression (Sn) = 340

First term of AP (a) = 7

Common difference of AP (d) = 6

Find;

Number of term in AP (N)

Computation:

Sn = [n/2][2a + (n-1)d]

340 = [n/2][2(7) + (n-1)6]

340 = [n/2][14 + 6n - 6]

680 = n[6n + 8]

6n² + 8n - 680

Using Quadratic Formula

n = 10

Number of term in AP (N) = 10

Learn more:

https://brainly.com/question/21859759?referrer=searchResults

Answer:

(a): The conditional pmf of Y when X = 1

[tex]p_{Y|X}(0|1) = 0.2353[/tex]

[tex]p_{Y|X}(1|1) = 0.5882[/tex]

[tex]p_{Y|X}(2|1) = 0.1765[/tex]

(b): The conditional pmf of Y when X = 2

[tex]p_{Y|X}(0|2) = 0.0962[/tex]

[tex]p_{Y|X}(1|2) = 0.2692[/tex]

[tex]p_{Y|X}(2|2) = 0.6346[/tex]

(c): From (b) calculate P(Y<=1 | X =2)

[tex]P(Y\le1 | X =2) = 0.3654[/tex]

(d): The conditional pmf of X when Y = 2

[tex]p_{X|Y}(0|2) = 0.025[/tex]

[tex]p_{X|Y}(1|2) = 0.150[/tex]

[tex]p_{X|Y}(2|2) = 0.825[/tex]

Step-by-step explanation:

Given

The above table

Solving (a): The conditional pmf of Y when X = 1

This implies that we calculate

[tex]p_{Y|X}(0|1), p_{Y|X}(1|1), p_{Y|X}(2|1)[/tex]

So, we have:

[tex]p_{Y|X}(0|1) = \frac{p(y = 0\ n\ x = 1)}{p(x = 1)}[/tex]

Reading the data from the given table, the equation becomes

[tex]p_{Y|X}(0|1) = \frac{0.08}{0.08+0.20+0.06}[/tex]

[tex]p_{Y|X}(0|1) = \frac{0.08}{0.34}[/tex]

[tex]p_{Y|X}(0|1) = 0.2353[/tex]

Using the format of the above formula for the rest, we have:

[tex]p_{Y|X}(1|1) = \frac{0.20}{0.34}[/tex]

[tex]p_{Y|X}(1|1) = 0.5882[/tex]

[tex]p_{Y|X}(2|1) = \frac{0.06}{0.34}[/tex]

[tex]p_{Y|X}(2|1) = 0.1765[/tex]

Solving (b): The conditional pmf of Y when X = 2

This implies that we calculate

[tex]p_{Y|X}(0|2), p_{Y|X}(1|2), p_{Y|X}(2|2)[/tex]

So, we have:

[tex]p_{Y|X}(0|2) = \frac{p(y = 0\ n\ x = 2)}{p(x = 2)}[/tex]

Reading the data from the given table, the equation becomes

[tex]p_{Y|X}(0|2) = \frac{0.05}{0.05+0.14+0.33}[/tex]

[tex]p_{Y|X}(0|2) = \frac{0.05}{0.52}[/tex]

[tex]p_{Y|X}(0|2) = 0.0962[/tex]

Using the format of the above formula for the rest, we have:

[tex]p_{Y|X}(1|2) = \frac{0.14}{0.52}[/tex]

[tex]p_{Y|X}(1|2) = 0.2692[/tex]

[tex]p_{Y|X}(2|2) = \frac{0.33}{0.52}[/tex]

[tex]p_{Y|X}(2|2) = 0.6346[/tex]

Solving (c): From (b) calculate P(Y<=1 | X =2)

To do this, where Y = 0 or 1

So, we have:

[tex]P(Y\le1 | X =2) = P_{Y|X}(0|2) + P_{Y|X}(1|2)[/tex]

[tex]P(Y\le1 | X =2) = 0.0962 + 0.2692[/tex]

[tex]P(Y\le1 | X =2) = 0.3654[/tex]

Solving (d): The conditional pmf of X when Y = 2

This implies that we calculate

[tex]p_{X|Y}(0|2), p_{X|Y}(1|2), p_{X|Y}(2|2)[/tex]

So, we have:

[tex]p_{X|Y}(0|2) = \frac{p(x = 0\ n\ y = 2)}{p(y = 2)}[/tex]

Reading the data from the given table, the equation becomes

[tex]p_{X|Y}(0|2) = \frac{0.01}{0.01+0.06+0.33}[/tex]

[tex]p_{X|Y}(0|2) = \frac{0.01}{0.40}[/tex]

[tex]p_{X|Y}(0|2) = 0.025[/tex]

Using the format of the above formula for the rest, we have:

[tex]p_{X|Y}(1|2) = \frac{0.06}{0.40}[/tex]

[tex]p_{X|Y}(1|2) = 0.150[/tex]

[tex]p_{X|Y}(2|2) = \frac{0.33}{0.40}[/tex]

[tex]p_{X|Y}(2|2) = 0.825[/tex]

Answer:

SAA

Step-by-step explanation:

HOPE IT HELPS YOU IN YOUR LEARNING PROCESS.

step by step explanation:

[tex]\mathfrak{x}^{2}+{y}^{2}+16=0[/tex]

=[x2+16=0x26]

=[2x{y}^2{16}~0]

=[4×{y}^0{16}]

=[32x{y}^x]

Answer:

[tex]\frac{y^{6} }{ x^{2} }[/tex]

Step-by-step explanation:

[tex]y^{6} x^{-2}[/tex]

Answer and Step-by-step explanation:

When there is a set of values within parenthesis, and an exponents on the values and on the parenthesis, you multiply the outer exponent with the inner exponent.

When a value has a negative exponent, the value that has the negative exponent will become a fraction and go to the denominator of the fraction (or go immediately to the denominator), or if it is already a fraction, goes to the denominator. If the value that has the negative exponent is in the denominator, the value will go to the numerator. In both instances, the negative exponent will then change to positive.

First, we need to simplify the expression inside the parenthesis.

[tex]y^{\frac{3}{2} } x^{-\frac{1}{2} } --> \frac{y^{\frac{3}{2} } }{x^{\frac{1}{2}} }[/tex]

Now we multiply the 4 to the exponents.

[tex]\frac{y^{\frac{3}{2} *\frac{x4}{1} } }{x^{\frac{1}{2}}*\frac{4}{1} } = \frac{y^{\frac{12}{2}} }{x^{\frac{4}{2}}} = \frac{y^6}{x^2}[/tex]

[tex]\frac{y^6}{x^2}[/tex] is the answer.

#teamtrees #PAW (Plant And Water)

Answer:

please write in english i cannot understand

Step-by-step explanation:

Answer:

[tex]m = \frac{1}{12}[/tex]

Step-by-step explanation:

Given

[tex](x,y) = (36,6)[/tex]

[tex]f(x) = \sqrt x[/tex] ----- the equation of the curve

Required

The slope of f(x)

The slope (m) is calculated using:

[tex]m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}[/tex]

[tex](x,y) = (36,6)[/tex] implies that:

[tex]a = 36; f(a) = 6[/tex]

So, we have:

[tex]m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}[/tex]

[tex]m = \lim_{h \to 0} \frac{f(36 + h) - 6}{h}[/tex]

If [tex]f(x) = \sqrt x[/tex]; then:

[tex]f(36 + h) = \sqrt{36 + h}[/tex]

So, we have:

[tex]m = \lim_{h \to 0} \frac{\sqrt{36 + h} - 6}{h}[/tex]

Multiply by: [tex]\sqrt{36 + h} + 6[/tex]

[tex]m = \lim_{h \to 0} \frac{(\sqrt{36 + h} - 6)(\sqrt{36 + h} + 6)}{h(\sqrt{36 + h} + 6)}[/tex]

Expand the numerator

[tex]m = \lim_{h \to 0} \frac{36 + h - 36}{h(\sqrt{36 + h} + 6)}[/tex]

Collect like terms

[tex]m = \lim_{h \to 0} \frac{36 - 36+ h }{h(\sqrt{36 + h} + 6)}[/tex]

[tex]m = \lim_{h \to 0} \frac{h }{h(\sqrt{36 + h} + 6)}[/tex]

Cancel out h

[tex]m = \lim_{h \to 0} \frac{1}{\sqrt{36 + h} + 6}[/tex]

[tex]h \to 0[/tex] implies that we substitute 0 for h;

So, we have:

[tex]m = \frac{1}{\sqrt{36 + 0} + 6}[/tex]

[tex]m = \frac{1}{\sqrt{36} + 6}[/tex]

[tex]m = \frac{1}{6 + 6}[/tex]

[tex]m = \frac{1}{12}[/tex]

Hence, the slope is 1/12