a rectangle has the height of n^3 + 4n^2 and the width of n^3 + 5n^2

Answer:

Bisector

Step-by-step explanation:

Answer:

the distance is called the circumference

Answer:

a = -1  b=0  and c=1

Step-by-step explanation:

-x^2 +0x+1

This is in the form

ax^2 +bx+c

a = -1  b=0  and c=1

Answer:

4x² - 10x - 6

Step-by-step explanation:

(x-3)(4x+2)

4x² + 2x - 12x - 6

4x² - 10x - 6

Answer: 7

Step-by-step explanation:

find x

6x(6+4)=5x(1+x)

Then x=11

AB= 11-4=7

Answer:

B.) 7

Step-by-step explanation:

I got it correct on founders edtell

Answer:

-2

Step-by-step explanation:

(1 , 3) = (x1 , y1)

(-2 , 9) = (x2 , y2)

slope = y2 - y2/x2 - x1

=9 - 3/-2 -1

=6/-3

=-2

Answer:

[tex]\int\limits {\int\limits_R {7(x + y)e^{x^2 - y^2}} \, dA = \frac{1}{2}e^{42} -\frac{43}{2}[/tex]

Step-by-step explanation:

Given

[tex]x - y = 0[/tex]

[tex]x - y = 7[/tex]

[tex]x + y = 0[/tex]

[tex]x + y = 6[/tex]

Required

Evaluate [tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA[/tex]

Let:

[tex]u=x+y[/tex]

[tex]v =x - y[/tex]

Add both equations

[tex]2x = u + v[/tex]

[tex]x = \frac{u+v}{2}[/tex]

Subtract both equations

[tex]2y = u-v[/tex]

[tex]y = \frac{u-v}{2}[/tex]

So:

[tex]x = \frac{u+v}{2}[/tex]

[tex]y = \frac{u-v}{2}[/tex]

R is defined by the following boundaries:

[tex]0 \le u \le 6[/tex]  ,  [tex]0 \le v \le 7[/tex]

[tex]u=x+y[/tex]

[tex]\frac{du}{dx} = 1[/tex]

[tex]\frac{du}{dy} = 1[/tex]

[tex]v =x - y[/tex]

[tex]\frac{dv}{dx} = 1[/tex]

[tex]\frac{dv}{dy} = -1[/tex]

So, we can not set up Jacobian

[tex]j(x,y) =\left[\begin{array}{cc}{\frac{du}{dx}}&{\frac{du}{dy}}\\{\frac{dv}{dx}}&{\frac{dv}{dy}}\end{array}\right][/tex]

This gives:

[tex]j(x,y) =\left[\begin{array}{cc}{1&1\\1&-1\end{array}\right][/tex]

Calculate the determinant

[tex]det\ j = 1 * -1 - 1 * -1[/tex]

[tex]det\ j = -1-1[/tex]

[tex]det\ j = -2[/tex]

Now the integral can be evaluated:

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA[/tex] becomes:

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \int\limits^6_0 {\int\limits^7_0 {7ue^{x^2 - y^2}} \, *\frac{1}{|det\ j|} * dv\ du[/tex]

[tex]x^2 - y^2 = (x + y)(x-y)[/tex]

[tex]x^2 - y^2 = uv[/tex]

So:

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \int\limits^6_0 {\int\limits^7_0 {7ue^{uv}} *\frac{1}{|det\ j|}\, dv\ du[/tex]

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \int\limits^6_0 {\int\limits^7_0 {7ue^{uv}} *|\frac{1}{-2}|\, dv\ du[/tex]

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \int\limits^6_0 {\int\limits^7_0 {7ue^{uv}} *\frac{1}{2}\, dv\ du[/tex]

Remove constants

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2}\int\limits^6_0 {\int\limits^7_0 {ue^{uv}} \, dv\ du[/tex]

Integrate v

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2}\int\limits^6_0 \frac{1}{u} * {ue^{uv}} |\limits^7_0 du[/tex]

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2}\int\limits^6_0 e^{uv} |\limits^7_0 du[/tex]

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2}\int\limits^6_0 [e^{u*7} - e^{u*0}]du[/tex]

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2}\int\limits^6_0 [e^{7u} - 1]du[/tex]

Integrate u

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2} * [\frac{1}{7}e^{7u} - u]|\limits^6_0[/tex]

Expand

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2} * ([\frac{1}{7}e^{7*6} - 6) -(\frac{1}{7}e^{7*0} - 0)][/tex]

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2} * ([\frac{1}{7}e^{7*6} - 6) -\frac{1}{7}][/tex]

Open bracket

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2} * [\frac{1}{7}e^{7*6} - 6 -\frac{1}{7}][/tex]

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2} * [\frac{1}{7}e^{7*6} -\frac{43}{7}][/tex]

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2} * [\frac{1}{7}e^{42} -\frac{43}{7}][/tex]

Expand

[tex]\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{1}{2}e^{42} -\frac{43}{2}[/tex]

Answer:

11-

3 / 8 * 5 / 5 = 15 / 40

2 / 5 * 8 / 8 = 16 / 40

16 / 45 + 15 / 45 =

31 / 40

12-

5 / 6 * 9 /9 = 45 / 54

4 / 9 * 6 / 6 = 24 / 54

24 / 54 + 45 + 54 = 69 / 54

69 / 54 =

1   15 / 54

Step-by-step explanation:

A shaded right rectangular pyramid is shown. where is the shade rectangular pyramid....how do I know the answer when you did post the full

question

Answer:

Length = 35

w = 5

Step-by-step explanation:

l x w = area

7y x y = 175

7y^2 = 175

y^2 = 175/7

y^2=25

y=5

The length is 7y, so it's 5 times 7 or 35.

The width is just y, so it's 5.

Answer:

x   -   h(x)

0   -   1

1    -   3

2   -   5

3   -   7

Step-by-step explanation:

Just substitute the x value into the function

Answer:

Step-by-step explanation:

Hope this helps u !!

The LCM of 2a(4a² + 10a + 13)(4a² + 13 - 10a).

Least Common Multiple, often known as the Lowest Common Multiple, is denoted by the acronym LCM. The lowest number that may be divided by both numbers is known as the least common multiple (LCM) of two numbers. It can also be computed using two or more numbers.

The smallest number among all the common multiples of the provided numbers is the least common multiple of two or more numbers.

We have,

16[tex]a^4[/tex] + 4a² + 169 and 8a³ +2a(2a + 13) + 16a²

First, 16[tex]a^4[/tex] + 4a² + 169

= (4a²)² + (2a)² + (13)²

= (4a²)² + 2 . 4a² . 13 + (13)² + 4a² + 104a²

= (4a² + 13)² - (10a)²

= (4a + 13 - 10a) (4a + 13 + 10a)

= (13 - 6a)(13 + 14a)

Second, 8a³ +2a(2a + 13) + 16a²

= 8a³ + 4a² + 26a + 16a²

= 8a³ + 20a² + 26a

= 2a(4a² + 10a + 13)

So, the LCM is

= 2a(4a² + 10a + 13)(4a² + 13 - 10a)

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Answer: -64 Maybe

Step-by-step explanation:

It look like each term goes up by double of the last term.

Answer:

6X(5 + 4)

6x(9)

54x

Answer:

the answer is 54x hope it will help full to you

Answer:

40cm

Step-by-step explanation:

Given data

Length= 15cm

Width=30cm

Volume= 18liters

1000cm^3= 18liters

18000cm^3= 18liters

We know that volume

Volume= L*W*h

18000= 15*30*h

18000=450h

h= 18000/450

h= 40cm

Hence the height is 40cm

Answer:

B: 112.5

Step-by-step explanation:

first, we split up the rhombus into two triangles, with the height of 15 and the length of 7.5. when you multiply these you will get 112.5, and normally if you are just getting the area of the triangle, then you would divide it by two, but seeing that we have 2 triangles here, we would multiply it by two afterwards so an easier way to do it is just to not divide.

There for the diameter is 2(4.8 )= 9.6 ft

Given: Volume of the cone is 83.7 m³

We know that:

[tex]\bigstar \ \ \boxed{\sf{\textsf{Volume of cone is given by} : \dfrac{\pi r^2h}{3}}}[/tex]

where r is the radius of the cone and h is the height of the cone

Given: Height of the cone = 5m

Substituting the values in the formula, we get:

[tex]\sf{\implies \dfrac{\pi r^2(5)}{3} = 83.7}[/tex]

[tex]\sf{\implies \pi r^2 = 83.7 \times \dfrac{3}{5}}[/tex]

[tex]\sf{\implies \pi r^2 = 83.7 \times 0.6}[/tex]

[tex]\sf{\implies \pi r^2 = 50.22}[/tex]

[tex]\sf{\implies r^2 = \dfrac{50.22}{\pi}}[/tex]

[tex]\sf{\implies r^2 = 16}[/tex]

[tex]\sf{\implies r = 4}[/tex]

We know that : Diameter is two times the radius

⇒  Diameter of the Cone is 8 meters

Answer:

5

Step-by-step explanation:

3*4=12

4*3=12

6*2=12

but 12 is not factor of 5

Answer:

3.75

Step-by-step explanation:

calculator because yes:)

Answer:

The other exterior angles must be 145 degrees. There is only one answer because there cannot be 2 of 110 degree angles in a triangle, so the other two exterior angles must correspond to the 2 equal angles in the triangle.

Step-by-step explanation:

For an exterior angle, we know that an exterior angle added with its corresponding interior angle is equal to 180 degrees. Thus, 180 = 70 + i, or the interior angle. i is then 110 degrees.

An isosceles triangle has two equal angles. As a triangle can only be 180 degrees. we cannot have 2 of i, as doing so would be equal to 220 degrees. Therefore, for the other two interior angles, they must be equal. Representing o, their angles could be put into an equation of 110 + 2o = 180, so o = 35.

For the other exterior angles, their corresponding interior angles are equal to 35 each, so they must be 145 degrees. There is only one answer because there cannot be 2 of 110 degree angles in a triangle, so the other two exterior angles must correspond to the 2 equal angles in the triangle.

Answer:

5. a) 5b

   b) 3b + 2b

Step-by-step explanation:

a) b + b + b + b + b is the same as

5 * b or 5b

b) 5b = 3b + 2b

or 3 * b + 2* b = 5b

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

Answer:

2 m

Step-by-step explanation:

The model of the situation is a right triangle with legs 1.5 and h the height of the wall. The hypotenuse is the ladder.

Using Pythagoras' identity

h² + 1.5² = 2,5²

h² + 2.25 = 6.25 ( subtract 2.25 from both sides )

h² = 4 ( take the square root of both sides )

h = [tex]\sqrt{4}[/tex] = 2

The height of the wall is 2 m

The height of the wall od the building is, 2 m.

In a right-angled triangle, the sum of the squares of the smaller two sides of a right-angle triangle is equal to the square of the largest side.

Given, A 2.5m long ladder leans against the wall of a building. The base of the ladder is 1.5m away from the wall.

So, Hypotenuse is 2.5 m and Adjacent is 1.5 m.

We know, H² = O² + A².

O² = H² - A².

O² = 6.25 - 2.25.

O² = 4.

O = 2.

Another concept regarding triangles is,

Suppose we have a triangle ABC, with sides a, b and c.

An acute angle triangle is one where a² + b² > c².

It is a right-angle triangle if a² + b² = c2.

It is an obtuse angle triangle if a² + b² < c².

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Answer:

Step-by-step explanation:

The answer is B.

It's a line that shows every point that satisfies the equation

y = 3x + 4  which is what I think you meant (but I'm not sure). If I am correct then there are a million possible points that could be the answer to this question.

If I am not correct, leave a comment that tells me so, and I'll revise my answer.

Never mind the question has the right equation. And my answer remains as given.

Given:

Sum of all the interior angles of a regular polygon is 1080°.

Measure of each side is 10 cm.

To find:

The perimeter.

Solution:

The sum of all interior angles of a regular polygon with n side is:

[tex]S=(n-2)180^\circ[/tex]

Sum of all the interior angles of a regular polygon is 1080°.

[tex]1080^\circ=(n-2)180^\circ[/tex]

[tex]\dfrac{1080^\circ}{180^\circ}=n-2[/tex]

[tex]6+2=n[/tex]

[tex]8=n[/tex]

Number of sides of the regular polygon is 8. The measure of each side is 10 cm. So, the perimeter of the regular polygon is:

[tex]\text{Perimeter}=\text{Number of sides}\times\text{Measure of each side}[/tex]

[tex]\text{Perimeter}=8\times 10[/tex]

[tex]\text{Perimeter}=80[/tex]

Therefore, the perimeter of the regular polygon is 80.

Answer:

The 4th graph.

Step-by-step explanation:

Its a horizontal line which indicates that x is known and the line begins over 2.

The correct Graph of x >2 has been given which is the 4th option.

An equation is an expression that shows the relationship between two or more numbers and variables.

We have given an equation as x> 2

Clearly, we can see the graph of x> 2 must be an area beyond 2 on the positive X -axis

The points (and only those points) which lie on the graph of the function should satisfy its equation.

Thus, if a point lies on the graph of a function, then it must also satisfy the given equation.

It is a horizontal line that indicates that the x value is known and the line begins over 2.

Hence, The correct Graph of x >2 has been given which is the 4th option.

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Answer:

-3

Step-by-step explanation:

[tex]\frac{7-25}{-2(-3)}[/tex]

=[tex]\frac{-18}{-2*-3}[/tex]

=[tex]\frac{-18}{6}[/tex]

=-3

[tex]\quad\quad\quad\quad\huge\tt{⟹\frac{7 - 25}{ - 2( - 3)} }[/tex]

[tex]\quad\quad\quad\quad\huge\tt{ ⟹\frac{7 - 25}{ - 2( - 3)} }[/tex]

[tex]\quad\quad\quad\quad\huge\tt{ ⟹\frac{-18}{ - 2( - 3)} }[/tex]

[tex]\quad\quad\quad\quad\huge\tt{ ⟹\frac{-18}{ 6} }[/tex]

[tex]\quad\quad\quad\quad\huge\tt{ ⟹ - 3} [/tex]

[tex]\quad\quad\quad\quad\huge \boxed{\tt{ \color{green} - 3} }[/tex]

________

#LetsStudy

Good question. It is supposed to be both the y-intercept and the x-intercept and is the only point on a graph that can be both. Something in wrong in that thing.