Helppppp!!!!!! Find m<1

Answer:

see below

Step-by-step explanation:

2: letter at the vertex,  placing a letter or number inside .... , or by using 3 letters

3:

WXY

WXZ

YXZ

4: X

Answer:

Decreases

Step-by-step explanation:

A negative correlation means the y is big when the x is small. As x gets bigger, y gets smaller.

Answer:

4 hours

Step-by-step explanation:

First we need to subtract the one pay fee (the helmet rental fee, $15)

So 51 - 15 = 36

Then we need to divide 36 by the amount of money it costs for each hours to find how many hours he rented the bike

So 39/9 = 4

Answer:

[tex]x \approx 2.740646096[/tex]

Step-by-step explanation:

Given:

[tex]x\log_{10}x=1.2[/tex]

Rewrite so that it's equal to zero:

[tex]\implies x\log_{10}x-1.2=0[/tex]

Therefore:

[tex]\implies f(x)=x\log_{10}(x)-1.2[/tex]

Differentiate the function:

Use the product rule to differentiate [tex]x\log_{10}(x)[/tex]:

[tex]\textsf{Let }u=x \implies \dfrac{du}{dx}=1[/tex]

[tex]\textsf{Let }v=\log_{10}(x) \implies \dfrac{dv}{dx}=\dfrac{1}{x \ln 10}[/tex]

[tex]\begin{aligned}\implies \dfrac{dy}{dx} & = u\dfrac{dv}{dx}+v \dfrac{du}{dx}\\& = \dfrac{x}{x \ln 10}+\log_{10}(x)\\& = \dfrac{1}{\ln 10}+\log_{10}(x)\end{aligned}[/tex]

[tex]\implies f'(x)=\dfrac{1}{\ln 10}+\log_{10}(x)[/tex]

[tex]\implies f'(x)=\dfrac{1+\ln 10 \log_{10}(x)}{\ln 10}[/tex]

Newton-Rhapson iteration formula

[tex]x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}[/tex]

Substitute the function and its derivative into the formula, replacing x with [tex]x_n[/tex]:

[tex]\implies x_{n+1}=x_n-\dfrac{x_n\log_{10}(x_n)-1.2}{\dfrac{1+\ln 10 \log_{10}(x)}{\ln 10}}[/tex]

[tex]\implies x_{n+1}=x_n-\dfrac{x_n\ln 10\log_{10}(x_n)-1.2\ln 10}{1+\ln 10 \log_{10}(x_n)}[/tex]

Therefore, the iteration formula is:

[tex]x_{n+1}=x_n-\dfrac{x_n\ln 10\log_{10}(x_n)-1.2\ln 10}{1+\ln 10 \log_{10}(x_n)}[/tex]

Let [tex]x_0=3[/tex]

Substitute this into the formula and carry out 3 iterations:

[tex]\implies x_{1}=3-\dfrac{3\ln 10\log_{10}(3)-1.2\ln 10}{1+\ln 10 \log_{10}(3)}=2.746149035[/tex]

[tex]\implies x_2=2.740648841[/tex]

[tex]\implies x_3= 2.740646096[/tex]

The solution of the equation -x^2 + 2x + 3 = x^2 - 2x + 3 are the x-coordinates of common points on the graphs are x=0 and x=2.

A system of equations is two or more equations that can be solved to get a unique solution. the power of the equation must be in one degree.

As we can see from the attached graph these graphs intersect at points (0,3) and (2,3).

The system of equation are;

[tex]y = x^2 + 2x + 3 \\y = x^2 - 2x + 3[/tex]

So,

[tex]-x^2 + 2x + 3 = x^2 - 2x + 3 \\\\2x^2 - 4x = 0\\\\2x ( x - 2) = 0[/tex]

Then the solutions of the equation are x=0 and x=2.

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Answer:

0 and 2

Step-by-step explanation:

b on edge

The two cars in relation to each other when they began travelling will be car B was 20 miles east of car A. Then the correct option is A.

The distance covered by the particle or the body in an hour is called speed. It is a scalar quantity. It is the ratio of distance to time.

We know that the speed formula

Speed = Distance/Time

Car A is travelling east at a steady speed of 50 miles per hour.

Then the equation of car A is given as

y = 50x + C

where x be the number of miles and x be the number of hours.

Then we have

After 2 hours, it is 140 miles east of Johns town.

Then we have

y = 50x ..1

Car B is travelling east on the same road. its distance east of Johns town is represented by the graph.

Then the equation of car B will be

y = [(120 - 20) / (2 - 0)]x + 20

y = 50x + 20 ...2

Then the two cars in relation to each other when they began travelling will be car B was 20 miles east of car A.

Then the correct option is A.

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Answer:
Car A was 20 miles east of car B.

Step-by-step explanation:

Answer: C

Step-by-step explanation:

The inverse of a function maps the domain onto the range.

Answer:

Θ = 39°

Step-by-step explanation:

OA = OB ( radii of the circle ) , then

Δ AOB is isosceles with base angles congruent , that is

∠ ABO = ∠ BAO = 52°

the exterior angle of a triangle is equal to the 2 opposite interior angles.

∠ DOB is an exterior angle of the triangle , so

∠ DOB = ∠ ABO + ∠ BAO = 52° + 52° = 114°

the angle between a tangent and the radius at the point of contact is 90°, then

∠ OBC = ∠ ODE = 90°

the sum of the 4 interior angles of quadrilateral OBCD = 360°

∠ OBC + ∠ DOB + ∠ ODC + ∠ BCD = 360°

90° + 114° + (90 + 27)° + Θ = 360°

90° + 114° + 117° + Θ = 360°

321° + Θ = 360° ( subtract 321° from both sides )

Θ = 39°

Answer:.

Step-by-step explanation:

Answer:

The values x = 7 and y = 3 make the equation true.

Step-by-step explanation:

hope this helps

The figure second represents the foundation drawing of a provided front, right, and top view option second is correct.

Three-dimensional objects can be represented in two dimensions via an orthographic projection.

As we have given orthographic drawings.

The front, right, and the top view is shown in the picture of the cuboid if merge all the view we will find the three-dimensional figure is a cuboid.

We can represent the cuboid by 2×3 grids as shown in the picture second.

Thus, the figure second represents the foundation drawing of a provided front, right, and top view option second is correct.

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Answer:

14

Step-by-step explanation:

First divide 24 and 6 to find how much 6's there are in 24.

24/6=4

Now, we take that number and multiply it by 3.5 to find how long it will take the group to walk 24km.

3.5*4=14.

It will take the group 14 hours to walk 24 km.

Hope this helps!

If not, I am sorry.

Answer:

x=10, y=14

Step-by-step explanation:

added in the picture

x = the quantity of 3 plus or minus the square root of 29 all over 2. Then the correct option is B.

The quadratic equation is given as ax² + bx + c = 0.

Then the solution is given as

[tex]\rm x = \dfrac{-b\pm \sqrt{b^2 - 4ac}}{2a}[/tex]

The quadratic equation is given below.

x² – 3x – 5 = 0

Then the solution of the equation will be

[tex]\rm x = \dfrac{-(-3)\pm \sqrt{(-3)^2 - 4 *1*(-5)}}{2*1}\\x = \dfrac{3 \pm \sqrt{29}}{2}[/tex]

Then the correct option is B.

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x = 2/3 , 0 .

Question :

[tex] {3x}^{2} - 2x = 0[/tex]

use quadratic formula :

[tex]x = \frac{( - b)± \sqrt{ {b}^{2} - 4ac } }{2a} [/tex]

Solution :

[tex]a {x}^{2} + bx + c = 0 \\ a = 3 \\ b = - 2 \\ c = 0[/tex]

[tex]x = \frac{ - ( - 2)± \sqrt{ { (- 2)}^{2} - 4 \times 3 \times 0} }{2 \times 3} \\ \\ x = \frac{2± \sqrt{4 - 0} }{6} \\ \\ x = \frac{2± \sqrt{4} }{6} \\ \\ x = \frac{2± 2}{6} [/tex]

separate into two equations :

[tex]x = \frac{2 + 2}{6} = \frac{4}{6} = \frac{2}{3} .[/tex]

[tex]x = \frac{2 - 2}{6} = \frac{0}{6} = 0.[/tex]

the values of x are 2/3 and 0 .

Answer:

P=3.5

P=2.667

Step-by-step explanation:

p=f/a

P=70/20

P=3.5

then f is increased by 10

f=70+10=80

and a is increased by 10

a=20+10

a=30

therefore; P=f/a

P=80/30

P=2.667

A number line from negative 3 to 3 in increments of 1. An open circle is at 5 and a bold line starts at 5 and is pointing to the right.

Here we have.

-3(2x - 5) < 5*(2 - x)

First, we need to isolate x in one side, let's expand both sides:

-6x + 15 < 10 - 5x

15 - 10 < 6x - 5x

5 < x

This will be represented with a number line where we have an open circle at 5, and an arrow that goes to the right. So the correct option is:

"A number line from negative 3 to 3 in increments of 1. An open circle is at 5 and a bold line starts at 5 and is pointing to the right."

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If BCD was reflected over the x axis then the coordinates will be B(-1,0) C(-1,2) and D(-4,1).

Given BCD is a triangle having coordinates (-1,0),(-1,-2),(-4,-1).

When the triangle is reflected over the x axis then all the values of y will change from negative to positive and the values of x remains same because in the second and third quadrant the values of x remains same and in the second quadrant the values of y used to  be positive  and in the third quadrant the values of y used to be negative.

So coordinates of B will change from (-1,0) to (-1,0)

coordinates of C will change from (-1,-2) to (-1,2)

coordinates of D will change from (-4,-1) to (-4,1)

New coordinates will be (-1,0) (-1,2) (-4,1).

Hence if BCD was reflected over x axis the new coordinates will be (-1,0) (-1,2) (-4,1).

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Answer:

Step-by-step explanation:

ore energy entered the Earth system than exited due to the greenhouse effect, solar energy trapping systems like solar panel, etc.

How CO₂ changed the energy in the Earth system due to many reasons:-

Carbon dioxide forces the Earth's energy budget out of balance by absorbing thermal infrared energy radiated by the surface. It absorbs thermal infrared energy with wavelengths in a part of the energy spectrum that other gases, such as water vapor, co₂, etc.

The greenhouse effect also increases energy in the earth's system.

An increase in the amount of carbon dioxide creates an overabundance of greenhouse gases that trap additional heat.

These greenhouse gas emissions have increased the greenhouse effect and caused the earth's surface temperature to rise.

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Answer:  [tex]x^2[/tex]  and  [tex]-2[/tex]

Step-by-step explanation:

original :

[tex](10x^3+3x^2)+(-20x-6)[/tex]

factored :

[tex]x^2(10x+3)+-2(10x+3)[/tex]

Answer :

[tex]x^2[/tex]  and  [tex]-2[/tex]

Answer:

[tex]\fbox {See below} \downarrow[/tex]

Step-by-step explanation:

Question 1 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(4x^{3})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 4(3)x^{3-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 12x^{2}}[/tex]

Question 2 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(\frac{2}{x^{4}})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(2x^{-4})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 2(-4)x^{-4-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = -8x^{-5}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = -\frac{8}{x^{5}}}[/tex]

Question 3 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(3\sqrt{x})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(3(x)^{\frac{1}{2}})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 3(\frac{1}{2})x^{\frac{1}{2}-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{3}{2}x^{-\frac{1}{2}}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{3}{2\sqrt{x}}}[/tex]

Question 4 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(\frac{3}{\sqrt{x}})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(3x^{-\frac{1}{2}})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 3(-\frac{1}{2})x^{-\frac{1}{2}-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = -\frac{3}{2}x^{-\frac{3}{2}}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = -\frac{3}{2x\sqrt{x}}}[/tex]

Question 5 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(\frac{x^{3}}{4})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{1}{4}(3)x^{3-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{3}{4}x^{2}}[/tex]

Question 6 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(3x^{5})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 3(5)x^{5-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 15x^{4}}[/tex]

Answer:

27/4

Step-by-step explanation:

added in the picture

The expected enrolment in 7 years is  1845.

The formula for calculating future value of the number of students is an exponential equation with this form:

FV = P (1 + r) ^n

1500 x (1.03)^7 = 1845

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Answer:

a

Step-by-step explanation:

The angle of depression from the top of the lighthouse to the sinking boat is 6.9°.

Trigonometry deals with the relationship between the sides and angles of a right-angle triangle.

From ΔABC it is obtained that;

Height of lighthouse, AC = 25 meters

Distance of boat from the lighthouse, BC = 204 meters

The angle of depression from the top of the lighthouse to the sinking boat is;

[tex]\rm tan \theta = \frac{P}{B} \\\\ \rm tan \theta = \frac{AC}{BC} \\\\ tan \theta = \frac{25}{204} \\\\ \theta = tan^{-1}(0.12)\\\\ \theta = 6.9 ^0[/tex]

Hence, the angle of depression from the top of the lighthouse to the sinking boat is 6.9°

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Answer:

Step-by-step explanation:

(√6 - √2)/4