Simplify the expression: –3(n − 5)

Answer:

A man walking away from a lamppost with a light source h=8 m above the ground. The man is m= 2 m tall. How long is the man’s shadow when he is d=15 m from the lamppost?

Step-by-step explanation:

2/x=6/(10+x) cross multiply.

6x=2(10+x)

6x=20+2x

6x-2x=20

4x=20

x=20/4

x=5 m. is the length of the man's shadow.

Answer:

g(h(6)) = 36

h(g(6)) = 36

Explanation:

There is no difference in this because we're just basically using the commutative property. For g(h(6)), you just plug in 6 for x² which will give you 36. And since g(x) = x, the value of x is 36. In the same way for h(g(6)), g(6) = 6 and h(6) = (6)² therefore getting you 36. For both equations though, you will get the same answer.

Answer:

g(h(6)) = 36

h(g(6)) = 36

Step-by-step explanation:

There is no difference in this because we're just basically using the commutative property. For g(h(6)), you just plug in 6 for x² which will give you 36. And since g(x) = x, the value of x is 36. In the same way for h(g(6)), g(6) = 6 and h(6) = (6)² therefore getting you 36. For both equations though, you will get the same answer.

Answer:

I think C

Step-by-step explanation:

It has a whole number as numerator and denominator so it is rational and it is a real number, but it is not an integer because it is a fraction.

Answer:

Rational, and Real numbers.

Step-by-step explanation:

There's a certain overlapping sequence of categorizing:

Whole #, Integers, Rational, Real.

Fractions and decimals belong in Rational, so you could say they are real as well.

Hope this helps! (sorry about it being brief)

Answer:

The perimeter of a square is 128, the area when perimeter 84 is 441

Step-by-step explanation:

When the length of a square arm is l,

the area of a square is A = l^2.

So in your case, A = \sqrt{1024} = l^2

So l is simply, l = 32.

The perimeter of a square is, P = 4.l = 32*4 = 128

Again when the perimeter is, P = 84

l = 84/4 = 21

So then the area is, A = 21^2 = 441

Answer:

4/6

Step-by-step explanation:

there are three even numbers (3/6) and 3 odd numbers but you are only looking for 5 (1/6) so the answer is 4/6

hopefully this helps you :)

pls mark my answer brainlest I need it ;)

Answer:

3.4 lessons per week

Step-by-step explanation:

If the teacher has to teach 6 chapter in 15 weeks, and each chapter has around 8.5 lessons. it means that the teacher has to teach in 15 weeks a total amount of lessons of 6*8.5= 51.

so 15 weeks---51 lessons

    1 week------x

From the above relation we have= 51/15= 3,4 lessons per week

Answer:

32/9

Step-by-step explanation:

Answer: 7/2 or 3  1/2

Step-by-step explanation:

Answer:

20,000

Step-by-step explanation:

Scince 4, (a number lower than 5) is in front of the 3 (which is in the 10,000's place), 20,000 should be the answer.

Answer:

4th option is the answer for first question.

4th option is the answee for second question.

Step-by-step explanation:

please mark me as a brainlist..

Answer:

Yes it does

Step-by-step explanation:

Distance is equal to rate multiplied by time and the rate is how fast the car is going and how long is the time so yes it will be equal to distance.

The statement given in the question is TRUE that is "the distance covered by a car depends on how fast the car was going and on how long it has travelled"

The formula for calculating distance is expressed as;

[tex]distance=speed \times time[/tex]

From the formula we can see that:

From the highlighted point, we can see that the statement given in the question is TRUE that is "the distance covered by a car depends on how fast the car was going and on how long it has travelled"

Learn more here: https://brainly.com/question/23774048

Answer: 3

Step-by-step explanation:

Answer:

Step-by-step explanation:

80

Answer:

Step-by-step explanation:

y - 3 = 0(x - 1)

y - 3 = 0

y = 3

Answer:

How can I help

Step-by-step explanation:

sir you need at least put something there because i can't really see it thank you no thank you

Answer:

7/9

Step-by-step explanation:

x = .7

10x = 7.7

10x - x = 7.7 repeating

9x = 7

x = 7/9

Bam

For a vector-valued function

[tex]\mathbf f(\mathbf x)=\mathbf f(x_1,x_2,\ldots,x_n)=(f_1(x_1,x_2,\ldots,x_n),\ldots,f_m(x_1,x_2,\ldots,x_n))[/tex]

the matrix of partial derivatives (a.k.a. the Jacobian) is the [tex]m\times n[/tex] matrix in which the [tex](i,j)[/tex]-th entry is the derivative of [tex]f_i[/tex] with respect to [tex]x_j[/tex]:

[tex]D\mathbf f(\mathbf x)=\begin{bmatrix}\dfrac{\partial f_1}{\partial x_1}&\dfrac{\partial f_1}{\partial x_2}&\cdots&\dfrac{\partial f_1}{\partial x_n}\\\dfrac{\partial f_2}{\partial x_1}&\dfrac{\partial f_2}{\partial x_2}&\cdots&\dfrac{\partial f_2}{\partial x_n}\\\vdots&\vdots&\ddots&\vdots\\\dfrac{\partial f_m}{\partial x_1}&\dfrac{\partial f_m}{\partial x_2}&\cdots&\dfrac{\partial f_n}{\partial x_n}\end{bmatrix}[/tex]

So we have

(a)

[tex]D f(x,y)=\begin{bmatrix}\dfrac{\partial(e^x)}{\partial x}&\dfrac{\partial(e^x)}{\partial y}\\\dfrac{\partial(\sin(xy))}{\partial x}&\dfrac{\partial(\sin(xy))}{\partial y}\end{bmatrix}=\begin{bmatrix}e^x&0\\y\cos(xy)&x\cos(xy)\end{bmatrix}[/tex]

(b)

[tex]D f(x,y,z)=\begin{bmatrix}\dfrac{\partial(x-y)}{\partial x}&\dfrac{\partial(x-y)}{\partial y}&\dfrac{\partial(x-y)}{\partial z}\\\dfrac{\partial(y+z)}{\partial x}&\dfrac{\partial(y+z)}{\partial y}&\dfrac{\partial(y+z)}{\partial z}\end{bmatrix}=\begin{bmatrix}1&-1&0\\0&1&1\end{bmatrix}[/tex]

(c)

[tex]Df(x,y)=\begin{bmatrix}y&x\\1&-1\\y&x\end{bmatrix}[/tex]

(d)

[tex]Df(x,y,z)=\begin{bmatrix}1&0&1\\0&1&0\\1&-1&0\end{bmatrix}[/tex]

Answer:

28

Step-by-step explanation:

Answer:

Number of chaperones = 9

Step-by-step explanation:

Given:

Total number of teacher = 2

Total people attend = 54

Find:

Number of chaperones

Computation:

Assume;

Number of chaperones = x

So,

x = [54-x] / 5

5x = 54 - x

6x = 54

x = 9

Number of chaperones = 9

Answer:

The points are "(2,2)".

Step-by-step explanation:

Given value:

[tex]f(x,y)=x^3-6xy+y^3+7[/tex]

Finding the first partial derivation which is equal to 0.

[tex]\to f_x(x,y)=3x^2-6y...............(a)\\\\ \to f_y(x,y)=-6x+3y^2...................(b)[/tex]

solve both the above equation by equal to 0.

For equation (a)

[tex]\to 3x^2-6y=0\\\\\to 3x^2=6y\\\\\to x^2=2y\\\\\to y=\frac{x^2}{2}\\\\[/tex]

For equation (b)

[tex]\to -6x+3y^2 =0 \\\\\to 3y^2 =6x\\\\\text{put the value of y in equation b}\\\\\to 3(\frac{x^2}{2})^2 -6x=0\\\\\to \frac{3}{4} x^4 -6x=0\\\\[/tex]

by solving the above value in the form of x we get:

[tex]\to x=0\\ \to x=2\\ \to x= -1 +\sqrt{3} i\\ \to x= -1 -\sqrt{3} i\\[/tex]

by solving the above value in the form of y we get:

[tex]\to y=0\\ \to y=2\\ \to y= -1 +\sqrt{3} i\\ \to y= -1 -\sqrt{3} i\\[/tex]

Solve the value by applying the second derivation method:

[tex]\to f_{xx}(x,y)=6x...............(a1)\\\\ \to f_{yy}(x,y)=6y...................(b2)\\\\\to f_{xy}(x,y)=-6...................(c2)[/tex]

calculating the value of discriminate:

[tex]d=f_{xx}(x,y) f_{yy}(x,y)-[f_{xy}(x,y)]^2[/tex]

The critical point of the given equation will be (2,2)

[tex]d=f_{xx}(2,2) f_{yy}(2,2)-[f_{xy}(2,2)]^2\\\\[/tex]

  [tex]= (6 \times 2) (6 \times 2) -[(-6)^2]\\\\= (12) (12) -[36]\\\\= 144 -36\\\\= 108\\\\[/tex]

Answer:

[tex](2,2)[/tex] is a minimum point of the function [tex]f(x,y).[/tex]

Step-by-step explanation:

[tex]f(x,y)=x^3-6xy+y^3+7[/tex]

Find the partial differentiation w.r.t. [tex]x[/tex] and [tex]y[/tex].

[tex]\frac {\partial f}{\partial x}=3x^2-6y[/tex]

[tex]\frac {\partial f}{\partial y}=-6x+3y^2[/tex]

[tex]\frac {\partial^2 f}{\partial x^2}=6x[/tex]

[tex]\frac {\partial^2 f}{\partial y^2}=6y[/tex]

[tex]\frac{\partial f}{\partial xy}=-6[/tex]

Find the critical points.

[tex]\frac {\partial f}{\partial x}=\frac {\partial f}{\partial y}=0[/tex]

[tex]\Rightarrow 3x^2-6y=0 \;\text{and}\; -6x+3y^2=0[/tex]

[tex]\Rightarrow 6y=3x^2\Rightarrow y=\frac{x^2}2[/tex]

[tex]\Rightarrow -6x+\frac 34 x^4=0\Rightarrow \frac {x^3}4=2[/tex]

[tex]\Rightarrow x^3=8\Rightarrow x=2,y=2[/tex]

Therefore, [tex](2,2)[/tex] is a critical point.

Now, [tex]\frac {\partial^2 f}{\partial x^2}\frac {\partial^2 f}{\partial y^2}-{\frac {\partial f}{\partial xy}}^2[/tex]

[tex]=6x6y-36=36xy-36>0 \;\text{at critical point }\; (2,2)[/tex]

Thus, [tex](2,2)[/tex] is not a saddle point.

[tex]\because \frac{\partial^2 f}{\partial x^2}>0 \; \text{and}\; \frac {\partial^2 f}{\partial x^2}\frac {\partial^2 f}{\partial y^2}-{\frac {\partial f}{\partial xy}}^2>0 \;\text{at point}\; (2,2)[/tex]

Hence, [tex](2,2)[/tex] is a local minimum of a given function.

11=5(x-2)+2x

11-2x=5(x-2)

(11-2x) / 5 = x-2

11/5=2.2

2.2-0.4x = x-2

4.2-0.4x=x

4.2=1.4x

x=3

Answer: x = 3

Step-by-step explanation:

5x -10 +2x = 11

7x = 21

x=3

Hope that helps!

Answer:

The answer is x-6

Step-by-step explanation:

x=(whatever number)

less means subtract

So the answer is x-6

Answer:

B is correct!!!!!!!!!   x-6

Step-by-step explanation:

this is correct because when it says 'than' that means the whole answer flips around

Answer:

[tex]x=3[/tex]

Step-by-step explanation:

So we have the equation:

[tex]2x+7+2x=19[/tex]

First, let's combine like terms on the left:

[tex](2x+2x)+7=19[/tex]

Add:

[tex]4x+7=19[/tex]

Subtract 7 from both sides:

[tex](4x+7)-7=(19)-7[/tex]

The left side cancels. Subtract on the right:

[tex]4x=12[/tex]

Now, divide both sides by 4:

[tex]\frac{4x}{4}=\frac{12}{4}[/tex]

The left cancels. Divide on the right:

[tex]x=3[/tex]

So, the value of x is 3.

Answer:

X=3

Step-by-step explanation:

2x+2x(=4x)=19-7

4x=12

X=12\4(divide)

x=3

Answer:

19

Step-by-step explanation:

f(x) when x=2 is just a fancy way of saying substitute the x in this equation with 2.

so the equation becomes..

3(2)^3-5    ... remeber PEMDAS

3(8)-5

24 -5 =19

Answer:6

Step-by-step explanation:

Answer:

7. C. 6

8. H. √34

9. A. (1, 3.5)

10. J. 10

Step-by-step explanation:

7. AB = 2y, BC = 6y, AC = 48

AB + BC = AC (segment addition theorem)

Substitute the above values into the equation

2y + 6y = 48

Solve for y

8y = 48

Divide both sides by 8

8y/8 = 48/8

y = 6

8. Distance between P(2, 8) and Q(5, 3):

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

Let,

[tex] P(2, 8) = (x_1, y_1) [/tex]

[tex] Q(5, 3) = (x_2, y_2) [/tex]

[tex] PQ = \sqrt{(5 - 2)^2 + (3 - 8)^2} [/tex]

[tex] PQ = \sqrt{(3)^2 + (-5)^2} [/tex]

[tex] PQ = \sqrt{9 + 25} [/tex]

[tex] PQ = \sqrt{34} [/tex]

9. Midpoint (M) of segment LB, for L(8, 5) and B(-6, 2) is given as:

[tex] M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) [/tex]

Let [tex] L(8, 5) = (x_1, y_1) [/tex]

[tex] B(-6, 2) = (x_2, y_2) [/tex]

Thus:

[tex] M(\frac{8 + (-6)}{2}, \frac{5 + 2}{2}) [/tex]

[tex] M(\frac{2}{2}, \frac{7}{2}) [/tex]

[tex] M(1, 3.5) [/tex]

10. M = -10, N = -20

Distance between M and N, MN = |-20 - (-10)|

= |-20 + 10| = |-10|

MN = 10

Area of a square = side^2

In this problem,

Side = 5ft

Plug 5 into the formula above

Area = 5^2 = 25ft^2

But we have to subtract the area of the circle.

Area of a circle = πr^2

r = radius

The diameter is 5 ft and the radius of a circle is diameter / 2

The radius of this circle = 2.5

Let's plug that into our formula.

Area = π(2.5)^2 = 19.63.

Let's subtract the area of the circle from the square

25 - 19.63 = 5.37 = 5.4ft^2

Answer:

I would choose A

Step-by-step explanation:

I chose A because it's an hypotenuse because of the different sides