b) The unit digit of a two-digit number is 1 less that the tens digit. If the number is increased by 8 and then divided by the sum of the digit, the result is 8. Find the number b ) The unit digit of a two - digit number is 1 less that the tens digit . If the number is increased by 8 and then divided by the sum of the digit , the result is 8. Find the number.​

Answer:

452/2 = 226

Step-by-step explanation:

45 divided by 2 is 22 and the remainder would be 12, which could be divided by 2 again to get 6. Add this to the total and you get 226.

Hope this helps! :D

The rule that describes the composition of transformations that maps figure PQRS to figure P"Q"R"S"is r y-axis • RO, 90° (x, y)

See attachment for complete question

From the image, we have:

P = (1,1)

Q = (1,5)

R = (3,5)

S = (3,1)

Start by reflecting PQRS across the y-axis. i.e. r y-axis

This is done by

(x, y) -> (-x,y)

So, we have:

P' = (-1,1)

Q' = (-1,5)

R' = (-3,5)

S' = (-3,1)

Next, we rotate by 90 degrees counterclockwise i.e. RO, 90° (x, y)

This is done by

(x, y) -> (-y, x)

So, we have:

P'' = (-1,-1)

Q'' = (-5,-1)

R'' = (-5,-3)

S'' = (-1,-3)

Hence, the rule that describes the composition of transformations that maps figure PQRS to figure P"Q"R"S"is r y-axis • RO, 90° (x, y)

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Answer:

okkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

will you be my friend

Step-by-step explanation:

.......... okkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

The z-score of an appliance that stopped working at 64 months is 2.

The mean of observations is equal to the ratio of sum of all the observations and the number of observations.

Given, the set of life spans of an appliance is normally distributed with a mean mu = 48 months and a standard deviation sigma = 8 months

The z-score is then calculated as

z = x - μ /σ

For an appliance that stopped working at 64 months, we have

x = 64

On substituting the values, we get

z = 64-48/8

z = 2

Hence, the z-score of an appliance that stopped working at 64 months is 2

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Answer:

D on edge

Step-by-step explanation:

2

See attached images for work.

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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Step-by-step explanation:

the area of a triangle is

baseline × height / 2

baseline and height must be at a right angle with each other.

so, in our case

6 × 11 / 2 = 3 × 11 = 33 cm²

Answer:

Area = 33 cm²

Step-by-step explanation:

[tex]\textsf{Area of a triangle}=\sf \dfrac{1}{2}bh[/tex]

where:

If we turn the triangle counterclockwise, so that the side that measures 6 cm is the base, we will see that the length that measures 11 cm is the height of the triangle (see attached).

Therefore:

Substitute the found values into the formula and solve for A:

[tex]\implies \sf A=\dfrac{1}{2}(6)(11)[/tex]

[tex]\implies \sf A=\dfrac{1}{2}(66)[/tex]

[tex]\implies \sf A=33[/tex]

Therefore, the area of the triangle is 33 cm²

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 .

The domain of the given graph is [−3, ∞) and the range is (−∞, 4].

We need to find the domain and range of the given graph.

The range of values that we are permitted to enter into our function is known as the domain of a function. The x values for a function like f make up this set (x). A function's range is the collection of values that it can take.

We can observe that the graph extends horizontally from −3 to the right without a bound, so the domain is [−3, ∞). The vertical extent of the graph is all range values 4 and below, so the range is (−∞, 4].

Therefore, the domain of the given graph is [−3, ∞) and the range is (−∞, 4].

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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 and Step-by-step explanation:

There is a cube with side length 3. The volume of this cube is 27.

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:

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: C

Step-by-step explanation:

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

Answer:

The answer is A

Step-by-step explanation:

They both have a maximum of 5!

The distance between the point (3, 5) and a line with the equation 5x – 3y + 10 = 0 will be 10 / √(34) units.

Let (x₁, y₁) be the point and ax + by + c = 0 be the line.

Then the distance between a point and a line will be

[tex]\rm d = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}[/tex]

The point (3, 5) and a line with the equation 5x – 3y + 10 = 0.

Then the distance between them will be

[tex]\rm d = \dfrac{|5 *3 - 3 *5 + 10|}{\sqrt{5^2 + (-3)^2}}\\d = \dfrac{10 }{\sqrt{34}} \ units[/tex]

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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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x-intercept -4.8

y-intercept 12

slope 2.5

plot points

x y

-2, 7

-1, 9.5

0, 12

1, 14.5

2 ,17

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:

7+45/n or 5

Step-by-step explanation:

45 divided by n or technically 5 is 9 plus 7 is 16 :)

Answer:7+45/n or 5

Step-by-step explanation:

Answer:

h = A / ( 1/2 a - 1/2 b)

Step-by-step explanation:

A = ½ ah - ½ bh

Factor out an h

A = h( 1/2 a - 1/2 b)

Divide each side by ( 1/2 a - 1/2 b)

A / ( 1/2 a - 1/2 b) = h ( 1/2 a - 1/2 b)/ ( 1/2 a - 1/2 b)

A / ( 1/2 a - 1/2 b) = h

h = A / ( 1/2 a - 1/2 b)

Answer:  2/6+5/6 + 1.167

Answer:

1.16

Step-by-step explanation:

Answer:

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

Step-by-step explanation:

hope this helps

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]