pls halllllllpppppppp

Answer:

2.5

Step-by-step explanation:

8x + 2 = 7 + 5x + 15

Combine like terms:

8x + 2 = 7 + 5x + 15

8x + 2 = 22

      -2     -2

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

8x = 20

----    ----

8       8

x = 2.5

Hope this helped.

Answer:

4 ther are 4 line symmetery

Answer:

two lines of symmetry

(a vertical and a horizontal)

Answer:

3(9t^5-7p^4)(9t^5+7p^4)

Step-by-step explanation:

243 t^10 - 147 p^8

3 ( 81 t^10-49 p^8 )

Then we need to factor what is in the parentheses

3 ( ( 9t^5) ^2 - ( 7p^4) ^2)

This is the difference of squares  ( a^2 -b^2) = ( a-b) (a+b)

3(9t^5-7p^4)(9t^5+7p^4)

Let

[tex]I(m) = \displaystyle \int_0^\pi x\sin^m(x)\,\mathrm dx[/tex]

Integrate by parts, taking

u = x   ==>   du = dx

dv = sinᵐ (x) dx   ==>   v = ∫ sinᵐ (x) dx

so that

[tex]I(m) = \displaystyle uv\bigg|_{x=0}^{x=\pi} - \int_0^\pi v\,\mathrm du = -\int_0^\pi \sin^m(x)\,\mathrm dx[/tex]

There is a well-known power reduction formula for this integral. If you want to derive it for yourself, consider the cases where m is even or where m is odd.

If m is even, then m = 2k for some integer k, and we have

[tex]\sin^m(x) = \sin^{2k}(x) = \left(\sin^2(x)\right)^k = \left(\dfrac{1-\cos(2x)}2\right)^k[/tex]

Expand the binomial, then use the half-angle identity

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

as needed. The resulting integral can get messy for large m (or k).

If m is odd, then m = 2k + 1 for some integer k, and so

[tex]\sin^m(x) = \sin(x)\sin^{2k}(x) = \sin(x)\left(\sin^2(x)\right)^k = \sin(x)\left(1-\cos^2(x)\right)^k[/tex]

and then substitute u = cos(x) and du = -sin(x) dx, so that

[tex]I(2k+1) = \displaystyle -\int_0^\pi\sin(x)\left(1-\cos^2(x)\right)^k = \int_1^{-1}(1-u^2)^k\,\mathrm du = -\int_{-1}^1(1-u^2)^k\,\mathrm du[/tex]

Expand the binomial, and so on.

Answer:

D. -4

Step-by-step explanation:

the slope formula is

m=(y2-y1)/(x2-x1)

(x2,y2) = (4,7)

(x1, y1) = (5,3)

So (7-3)/(4-5) = 4/-1 = -4

Answer:

-4

Step-by-step explanation:

I say so

Answer:

The answer is 216

Step-by-step explanation:

if there is a 2 cm border, that means that the sides will both become 2 centimeters longer. so (16+2)*(10*2) = 18*12 = 216.

Answer:

2

Step-by-step explanation:

The slope is given by

m = ( y2-y1)/(x2-x1)

   = (7-3)/(10-8)

    = 4/2

    = 2

Answer:

11. Yes

12. No

Step-by-step explanation:

11. The x values only have 1 y value, so it makes it a function.

12. The x value 0 has 2 y values -4 and 4, therefore it is not a function, because in functions that x value can only have one y value,

Now, it look like there is some information missing in the answer. The whole problem should look like this:

Alicia Keys's new album As I Am is climbing the charts, and the manager of Tip Top Tunes expects to sell a lot of copies. Because she has limited shelf space, she can't put out all her copies of the CD at once. On Monday morning, she stocked the display with 40 copies. By the end of the day, some of the copies had been sold. On Tuesday morning, she counted the number of copies left and then added that many more to the shelf. In other words, she doubled the number that was left in the display. At the end of the day, she discovered that she had sold the exact same number of copies as had been sold on Monday. On Wednesday morning, the manager decided to triple the number of copies that had been left in the case after Tuesday. Amazingly, she sold the same number of copies on Wednesday as she had on each of the first two days! But this time, at the end of the day the display case was empty. How many copies of the As I Am CD did she sell each day?

Answer:

She sold 24 copies of the cd each day.

Step-by-step explanation:

In order to solve this problem we must first set our variable up. In this case, since we need to know what the number of sold cd's per day is, that will just be our variable:

x= Number of copies sold.

So we can start setting our equation up. So we take the first part of the problem:

"On Monday morning, she stocked the display with 40 copies. By the end of the day, some of the copies had been sold."

This can be translated as:

40-x

where this expression represents the number of copies left on the shelf by the end of monday.

"On Tuesday morning, she counted the number of copies left and then added that many more to the shelf."

so we represent it like this:

(40-x)+(40-x)

"In other words, she doubled the number that was left in the display."

so the previous expression can be simplified like this:

2(40-x)

"At the end of the day, she discovered that she had sold the exact same number of copies as had been sold on Monday."

so the expression now turns to:

2(40-x)-x   this is the number of copies left by the end of tuesday.

"On Wednesday morning, the manager decided to triple the number of copies that had been left in the case after Tuesday."

this translates to:

3[2(40-x)-x]

This is the number of copies on the shelf by the begining of Wednesday.

"Amazingly, she sold the same number of copies on Wednesday as she had on each of the first two days! But this time, at the end of the day the display case was empty."

this piece of information lets us finish writting our equation:

3[2(40-x)-x] -x = 0

since there were no copies left on the shelf, then the equation is equal to zero.

So now we proceed and solve the equation for x:

3[2(40-x)-x] -x = 0

We simplify it from the inside to the outside.

3[80-2x-x]-x=0

3[80-3x]-x = 0

we now distribute the 3 so we get:

240-9x-x=0

we combine like terms so we get:

240-10x=0

we move the 240 to the other side of the equation so we get:

-10x=-240

and divide both sides into -10 so we get:

x=24

so she sold 24 copies each day.

Answer:

n = 9 is the answer.

Step-by-step explanation:

Given a Triangle [tex]\triangle KLM[/tex] with its perpendicular bisectors intersecting at a point A.

AK = 25 units and

AM = 3n -2

To find:

Value of n = ?

Solution:

First of all, let us learn about perpendicular bisectors and their intersection points.

Perpendicular bisector of a line PQ is the line which divides the line PQ into two equal halves and is makes an angle of [tex]\bold{90^\circ}[/tex] with the line PQ.

And in a triangle, the perpendicular bisectors of 3 sides meet at one point and that point is called Circumcenter of the triangle.

We can draw a circle from circumcenter so that the circle passes from the three vertices of the triangle.

i.e.

Circumcenter of a triangle is equidistant from all the three vertices of the triangle.

In the given statement, we are given that A is the circumcenter of the [tex]\triangle KLM[/tex].

Please refer to the attached image for the given triangle and sides.

The distance of A from all the three vertices will be same.

i.e. AK = AM

[tex]\Rightarrow 25 = 3n-2\\\Rightarrow 3n =25+2\\\Rightarrow 3n =27\\\Rightarrow \bold{n = 9}[/tex]

Therefore, n = 9 is the answer.

Answer:

D. infinitely many

Step-by-step explanation:

Perpendicular lines can be drawn everywhere on the lone in infinitely different places, making the answer infinite.

The number of lines that can be drawn perpendicular to a given line at a given point on that line in space is infinite many.

The correct answer is an option (D)

For given question,

We can draw infinite perpendicular lines to a

Therefore, the number of lines that can be drawn perpendicular to a given line at a given point on that line in space is infinite many.

The correct answer is an option (D)

Learn more about the perpendicular lines here:

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Answer:

Below

Step-by-step explanation:

Let T be triangle, C the circle and S the square.

● T + T + T = 30

● 3T = 30

Divide both sides by 3

● 3T/3 = 30/3

● T = 10

So the triangle has a value of 10.

●30 T + C + C = 20C + S + S = 13T +C ×S/2

Add like terms together

●30 T + 2C = 20C +2S= 13T + C×S/2

Replace T by its value (T=10)

● 300 + 2C = 20C + 2S = 130 + C×S/2

Take only this part 20C + 2S = 130 + C × S/2

● 20C + 2S = 130 + C×S/2 (1)

Take this part (300+2C = 20C+2S) and express S in function of C

● 20C + 2S = 300 + 2C

Divide everything by 2 to make easier

● 10 C + S = 150+ C

● S = 150+C-10C

● S = 150-9C

Replace S by (5-9C) in (1)

● 20C + 2S = 130 + C×S/2

● 20C + 2(150-9C) = 130 +C× (150-9C)/2

● 20C + 300-18C= 130 + C×(75-4.5C)

● 2C + 300 = 130 + 75 -4.5C^2

● 2C +300-130 = 75C - 4.5C^2

● 2C -75C + 170 = -4.5C^2

● -73C + 170 = -4.5C^2

Multiply all the expression by -1

● -4.5C^2 +73C+ 170= 0

This is a quadratic equation, so we will use the discriminant method.

Let Y be the discriminant

● Y = b^2-4ac

● b = 73

● a = -4.5

● c = 170

● Y = 73^2 - 4×(-4.5)×170= 8389

So the equation has two solutions:

● C = (-b +/- √Y) /2a

√Y is approximatively 92

● C = (-73 + / - 92 )/ -9

● C = 18.34 or C = -2.11

Approximatively

● C = 18 or C = -2

■■■■■■■■■■■■■■■■■■■■■■■■■

● if C = 18

30T + 2C = 300 + 36 = 336

● if C = -2

30T + 2C = 300-4 = 296

Answer: 125.6 in^2

Step-by-step explanation:

First, we have that the radius of this circle is r = 10in

Now, we know that the area of a circle is:

A = pi*r^2

Now, if we got only a section of the circle, defined by an angle x, then the area of that region is:

A = (x/360°)*pi*r^2

Notice that if x = 360°, then the area is the same as the area of the full circle, as expected.

Then each shaded area has an angle of 72°.

A = (72°/360°)*3.14*(10in)^2 = 62.8 in^2

And we have two of those, both of them with the same angle, so the total shaded area is:

2*A = 2*62.8 in^2 = 125.6 in^2

Split up C into two component paths C₁ and C₂, where each line segment is respectively parameterized by

r₁(t) = (1 - t ) (3i + k) + t (4i + 3j + k) = (t + 3) i + 3t j + k

r₂(t) = (1 - t ) (4i + 3j + k) + t (4i + 5j + 4k) = 4i + (2t + 3) j + (3t + 1) k

both with 0 ≤ t ≤ 1.

It's a bit unclear what function you're supposed to integrate (looks like xyz ?) so I'll give a more general result. The line integral of a scalar function f(x, y, z) along the given path C is

[tex]\displaystyle \int_C f(x,y,z)\,\mathrm ds = \int_{C_1}f(\mathbf r_1(t))\left\|\frac{\mathrm d\mathbf r_1}{\mathrm dt}\right\|\,\mathrm dt + \int_{C_1}f(\mathbf r_2(t))\left\|\frac{\mathrm d\mathbf r_2}{\mathrm dt}\right\|\,\mathrm dt[/tex]

We have

dr₁/dt = i + 3j   ==>   || dr₁/dt || = √(1² + 3²) = √10

dr₂/dt = 2j + 3k   ==>   || dr₂/dt || = √(2² + 3²) = √13

Then the integrals reduce to

[tex]\displaystyle \int_0^1 \left(\sqrt{10}\,f(t+3,3t,1) + \sqrt{13}\,f(4,2t+3,3t+1)\right)\,\mathrm dt[/tex]

If indeed f(x, y, z) = xyz, then we have

[tex]\displaystyle \int_0^1 \left(3\sqrt{10}\,t(t+3) + 4\sqrt{13}\,(2t+3)(3t+1)\right)\,\mathrm dt = 11\sqrt{\frac52}+42\sqrt{13}[/tex]

Answer:

d. 19

Step-by-step explanation:

Degrees of freedom is the number is the number of value which is used in the final calculation. It calculate as n-1, where n is the sample size. The degrees of freedom for the given scenario is 19. The sample size is 20 so the degrees of freedom is 1 less which will be 19.

Answer:

x/4 +1

Step-by-step explanation:

Answer:

Tran family's sprinkler was used for 20 hours

Green's  family's sprinkler was used for 30 hours

Step-by-step explanation:

Let the hours for which Tran family's sprinkler used is x hours

water output rate for the Tran family's sprinkler = 35L per hour

water output from  Tran family's sprinkler in x hours = 35*x L = 35x

Let the hours for which Green family's sprinkler used is y hours

water output rate for the Green family's sprinkler = 40L per hour

water output from  Green family's sprinkler in x hours = 40*y L = 40y

Given

The families used their sprinklers for a combined total of 50 hours

thus

x + y = 50 -------------------equation 1

y = 50-x

total water output of 1900L

35x+40y = 1900  -------------------equation 1

using  y = 50-x in equation 2, we have

35x + 40(50-x) = 1900

35x + 2000 - 40x = 1900

=> -5x = 1900 - 2000 = -100

=> x = -100/-5 = 20

y = 50-20 = 30

Thus,

Tran family's sprinkler was used for 20 hours

Green's  family's sprinkler was used for 30 hours

Answer:

Step-by-step explanation:

12x - 9 = 8x + 7

4x - 9 = 7

4x = 16

x = 4

solution is C

The solution is Option C.

The value of x is given from the equation x = 4

A perpendicular bisector is defined as a line or a line segment that divides a given line segment into two parts of equal measurement. Lines that cross each side's midpoint and are perpendicular to the specified side are known as a triangle's perpendicular bisectors.

The perpendicular bisector theorem states that any point on the perpendicular bisector is equidistant from both the endpoints of the line segment on which it is drawn

Given data ,

Let the first line be represented as CD

Let the second line be represented as XY

Now , CD is the perpendicular bisector of XY

So , the point F is the midpoint of the line segment XY

The measure of line segment XF = 12x - 9

The measure of line segment FY = 8x + 7

From the perpendicular bisector theorem ,

The measure of line segment XF = The measure of line segment FY

Substituting the values in the equation , we get

12x - 9 = 8x + 7

Subtracting 8x on both sides of the equation , we get

4x - 9 = 7

Adding 9 on both sides of the equation , we get

4x = 16

Divide by 4 on both sides of the equation , we get

x = 4

Therefore , the value of x = 4

Hence , the value of the equation is x = 4

To learn more about perpendicular bisectors click :

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Answer:

17

Step-by-step explanation:

[tex]h(x) =x^2 +1\\k(x)=x-2\\\\3h(2)+2k(3)\\\\h(2)= ?\\k(3)=?\\\\h(2) = (2)^2 +1\\= 4+1\\h(2)=5\\\\\\k(3)= 3-2\\k(3) = 1\\\\3h(2) +2k(3)\\\\= 3(5)+2(1)\\=15+2\\3h(2)+2k(3) = 17[/tex]

5, 9, and 17

Step-by-step explanation:

Answer:

sorry for doesn't know the answer

Answer:

45

62x

______

  90

2700+

_________

2790

Step-by-step explanation:

Answer:

204

Step-by-step explanation:

To simplify the shape, you can do multiple things. I've opted to shave down both prongs to take it from a 'T' shape to a rectangular prism.

For height of the prongs, take 4 from 6.

6 - 4 = 2

Divide by 2 as there are 2 prongs.

2 / 2 = 1

Remember L * W * H

6 * 3 * 1 = 18

Remember that there are two prongs!

3 + 4 = 7

6 * 7 * 4 = 168

168 + 2(18) = 204

We want to maximize [tex]x^3y^4z[/tex] subject to the constraint [tex]x+y+z=30[/tex].

The Lagrangian is

[tex]L(x,y,z,\lambda)=x^3y^4z-\lambda(x+y+z-30)[/tex]

with critical points where the derivatives vanish:

[tex]L_x=3x^2y^4z-\lambda=0[/tex]

[tex]L_y=4x^3y^3z-\lambda=0[/tex]

[tex]L_z=x^3y^4-\lambda=0[/tex]

[tex]L_\lambda=x+y+z-30=0[/tex]

[tex]\implies\lambda=3x^2y^4z=4x^3y^3z=x^3y^4[/tex]

We have

[tex]3x^2y^4z-4x^3y^3z=x^2y^3z(3y-4x)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}\\z=0,\text{ or}\\3y=4x\end{cases}[/tex]

[tex]3x^2y^4z-x^3y^4=x^2y^4(3z-x)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}\\3z=x\end{cases}[/tex]

[tex]4x^3y^3z-x^3y^4=x^3y^3(4z-y)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}4z=y\end{cases}[/tex]

Let's work with [tex]x=3z[/tex] and [tex]y=4z[/tex], for which we have

[tex]x+y+z=8z=30\implies z=\dfrac{15}4\implies\begin{cases}x=\frac{45}4\\y=15\end{cases}[/tex]

The smallest of these is C. 15/4.

Answer

20%

Step-by-step explanation:

Correct question is;

The claim is that the proportion of adults who smoked a cigarette in the past week is less than 0.35, and the sample statistics include n = 1168 subjects with 385 saying that they smoked a cigarette in the past week. Find the value of the test statistic

Answer:

Test statistic is z = -1.46

Step-by-step explanation:

Let's first of all define the hypotheses:

Null hypothesis:

H0: p = 0.35, i.e 35% in the sample of 1,168 adults have smoked cigarettes in the previous week.

Alternative hypothesis:

Ha: p < 0.35, i.e less than 35% in the sample of 1,168 adults have smoked cigarette in the previous week.

The sample size is, n = 1,168 while the number of adults who smoked in the previous week would be; x = 385

Therefore, the sample proportion of adults who smoked in the previous week would be calculated as;

p^ = x/n = 385/1168 ≈ 0.3296

Now, from Central Limit Theorem for large samples, The sampling distribution of the sample proportion p^, will have a mean of μ = p = 0.35

Formula for standard deviation is;

σ = √[p (1 – p)/n]

σ = √(0.35 × (1 – 0.35)/1168)

σ = √0.0001947774

σ = 0.014

Formula for test statistic is;

z = (p^ - p)/σ

z = (0.3296 - 0.35)/0.014

z = - 1.46

Answer:

6

Step-by-step explanation:

the | | are for absolute value, which means

|-6|=|6|= 6

Answer:

C. [tex] d = 15.81 units [/tex]

Step-by-step explanation:

Given:

2 end points on a graph => (5, 6) and (-4, -7)

Required:

Distance between them

SOLUTION:

Distance between two points in a graph can be calculated using [tex] distance (d) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]

Where,

[tex] (-4, -7) = (x_1, y_1) [/tex]

[tex] (5, 6) = (x_2, y_2) [/tex]

Plug in the values into the formula and solve

[tex] d = \sqrt{(5 - (-4))^2 + (6 - (-7))^2} [/tex]

[tex] d = \sqrt{(5 + 4))^2 + (6 + 7))^2} [/tex]

[tex] d = \sqrt{(9)^2 + (13)^2} [/tex]

[tex] d = \sqrt{81 + 169} [/tex]

[tex] d = \sqrt{250} [/tex]

[tex] d = 15.81 units [/tex]

Answer:

15.81

Step-by-step explanation:

If L(t) = ⟨5 + t, 1 + 5t⟩, then the tangent vector to L(t) is

dL/dt = ⟨1, 5⟩

Any line parallel to L(t) will have the same tangent vector, up to some scalar factor (that is, if the tangent vector is a multiple of ⟨1, 5⟩).

Any line r(t) with tangent vector T(t) = dr/dt that is perpendicular to L(t) will satisfy

T(t) • ⟨1, 5⟩ = 0

• r(t) = ⟨-5, -2t, 1 - 10t⟩ is parallel to L(t) because its tangent vector is

T(t) = ⟨-2, -10⟩ = -2 ⟨1, 5⟩

• r(t) = ⟨1 + 1.5t, 3 + 7.5t⟩ is parallel to L(t) because

T(t) = ⟨1.5, 7.5⟩ = 1.5 ⟨1, 5⟩

• r(t) = ⟨-2 - t, 2 - 2t⟩ is neither parallel nor perpendicular to L(t) because

T(t) = ⟨-1, -2⟩ ≠ k ⟨1, 5⟩

for any real k (in other words, there is no k such that -1 = k and -2 = 5k), and

⟨-1, -2⟩ • ⟨1, 5⟩ = -1 - 10 = -11 ≠ 0

• r(t) = ⟨3 + 15t, -3t⟩ is perpendicular to L(t) because

T(t) = ⟨15, -3⟩

and

⟨15, -3⟩ • ⟨1, 5⟩ = 15 - 15 = 0

Answer:

this is an inconsistent because no solutions

9514 1404 393

Answer:

  -13/11

Step-by-step explanation:

Straightforward evaluation of the expression at x=1 gives (1 -1)/(1 -1) = 0/0, an indeterminate form. So, L'Hopital's rule applies. The ratio of derivatives is ...

  [tex]\displaystyle\lim_{x\to 1}\dfrac{n}{d}=\dfrac{n'}{d'}=\left.\dfrac{\dfrac{4}{3\sqrt[3]{4x-3}}-\dfrac{7}{2\sqrt{7x-6}}}{\dfrac{5}{2\sqrt{5x-4}}-\dfrac{2}{3\sqrt[3]{2x-1}}}\right|_{x=1}=\dfrac{4/3-7/2}{5/2-2/3}=\dfrac{8-21}{15-4}\\\\=\boxed{-\dfrac{13}{11}}[/tex]