8m-40= -4(5m+3) What value of m makes the equation true?​

The amount raised by a booster club compared with the number of  raffle tickets sold shows a constant rate of change;

A constant rate of change is realized when a change in one situation always results in a change in another. In the scenario of a booster club that regularly sells raffle tickets, the more tickets it sells, the higher the club gets.

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The table with the scale measurements is given by the image shown at the end of the answer.

The measurements are obtained applying the proportion given for each table.

The symbols are given as follows:

For the first table, we have that every inch on the table represents one feet in real life, hence:

For the second table, we have that every inch on the paper represents two feet in real life, hence the measurements are given as follows:

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Answer: I Don't Know.

Step-by-step explanation: Add a picture.

Answer:

5

Step-by-step explanation:

Answer:

The radius of the circle is 5

Step-by-step explanation:

The statements that is true about the similarity of the two triangles the option D

D. ΔMNO and ΔJKL are not similar triangles

Similar triangles are triangles which have proportional corresponding sides

The parameters in the question are;

The length of segment MN = 20

Length of segment NO = 12

Length of segment OM = 25

Measure of angle ∠O = 56°

Length of segment LJ =- 15

Length of segment JK = 12

Length of segment KL = 9

Measure of angle ∠L = 56°

Two triangles are similar if the ratio of two sides on one triangle are proportional to two sides of another triangle, and the included  angle between the two sides are congruent

The included angle between sides ON and MO on triangle MNO is congruent to the included angle between segment LK and JL in triangle JKL

However, the ratio of the sides  LK to ON and JL to MO are;

9/12 ≠ 15/25

Therefore, the triangles are not similar

The correct option is option D

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The inequality that correctly describes the number of ounces of water, w, that tom drinks each day is d. 64<w.

Amount of water that the bottle can hold = 32 ounces

Bottles of water consumed = 2

A mathematical comparison and representation of the connection between two expressions is known as an inequality. It can be regarded as a generalisation of an equation and is denoted by signs such as <, > and =.

Tom consumes more than two complete bottles of water every day. Since each bottle can carry up to 32 ounces of water, we can describe Tom's daily water consumption as -

= w > 2 × 32

Therefore,

Simplifying the right side of the inequality:

w > 64

Complete Question:

Toms water bottle can hold up to 32 ounces of water. each day he drinks more than 2 full bottles of water. which inequality correctly describes the number of ounces of water, w, that tom drinks each day?

a. w>2

b. 32<w

c. w = 64

d. 64<w

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Solution:

Given:

An exponential function is of the form:

To get the exponential function for the relation;

Part A:

Relating this to the parameters given:

The exponential function that models the problem is;

Part B:

On the twenty-third day,

Therefore, he would have saved 4,194,304 pennies on the twenty-third day.

The possible change in volume (in mm3) resulting from the error in measuring the radius is 32. 06 mm³

Given,

A spherical golf ball is measured to have a radius of $$5 mm, with a possible measurement error of $$0.1 mm

How to determine the volume

The formula for volume of a sphere is expressed by;

Volume = 4/3 πr³

Where;

r is the radius of the sphere

pi takes the value 3.14

From the information given, we have that the measured radius is 5 mm, with a possible measurement error of 0.1 mm

Then, radius  a = 5mm

Radius b = 5 + 0. 1 = 5. 01mm

Now, substitute the values into the formula

Volume, v = 4/ 3 (3.14)(5)³

v = 4/ 3 (392.5)

v = 523. 3 mm³

For radius of 5.01mm

Volume =  4/ 3 (3.14)(5.1)³

expand the bracket

Volume = 4/3 (416.52)

Volume = 555. 4 mm³

The change in volume = 555. 4 - 523. 3 = 32. 06 mm³

Hence, the change in volume is 32. 06 mm³

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(A) Algorithms are mathematical formulas placed in software that performs an analysis on a data set.

Therefore, (A) algorithms are mathematical formulas placed in software that performs an analysis on a data set.

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Answer:

54 hope it helped you out

The correct solution to the equation (3x-4)/5 = 1 is x = 3. The given options indicate that x = 3 or x = –1/3. Here option B is the correct answer.

To solve the equation (3x-4)/5 = 1, we can follow these steps:

Multiply both sides of the equation by 5 to eliminate the fraction:

5 * [(3x-4)/5] = 5 * 1

Simplify:

3x - 4 = 5

Add 4 to both sides of the equation to isolate the term with x:

3x - 4 + 4 = 5 + 4

Simplify:

3x = 9

Divide both sides of the equation by 3 to solve for x:

(3x)/3 = 9/3

Simplify:

x = 3

Therefore, the solution to the equation (3x-4)/5 = 1 is x = 3.

So, the correct answer is B) x = 3 or x = –1/3.

The answer C) "No solutions" is incorrect since we found a solution. The answer D) "x = –5 or x = 1/5" is also incorrect as it doesn't satisfy the equation (3x-4)/5 = 1.

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No, the sample in this case is not random.

The sample in this case is not random. Random sampling involves selecting individuals from a population in such a way that each individual has an equal chance of being selected. In the given scenario, the sample consists only of students who buy hot lunch on a particular day.

This sampling method is not random because it introduces a bias by including only a specific subgroup of students who have chosen to buy hot lunch. It does not provide an equal opportunity for all students in the population to be selected for the survey.

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For the joint probability density function of x and y, which is given by f(x,y)=6/7(x² + xy/2); then the probability that P(x > y) is 15/56.

To find P(x > y), we need to integrate the joint probability density function f(x, y) over the region where x > y.

The joint probability density function of x and y is : f(x,y)=6/7(x² + xy/2); 0<x<1, 0<y<2;

The probability P(x>y) can be written as :

P(x > y) = ∫₀¹∫₀ˣ6/7(x² + xy/2)dx.dy;

P(x > y) = 6/7 × ∫₀¹(x³ + x³/4)dx;

P(x > y) = 6/7 × [x⁴/4 + x⁴/16]₀¹;

P(x > y) = 6/7 × [5x⁴/16]₀¹;

P(x > y) = 6/7 × (5/16) = 30/112 = 15/56.

Therefore, the required probability is 15/56.

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The given question is incomplete, the complete question is

The joint probability density function of x and y is given by f(x,y)=6/7(x² + xy/2); 0<x<1, 0<y<2

Find P(x > y).

Answer:

10 minutes.

Step-by-step explanation:

5 hours = 300 minutes

90 minutes + 105 minutes + 115 minutes = 310 minutes of programmes.

310-300 = 10 leftover minutes missed

Answer: 12x² - 5x - 2

Concept:

When doing questions that need to expand algebraic expressions, using the FOIL method will ease the process:

Solve:

Given expression

(3x - 2) (4x + 1)

Expand parentheses and apply the FOIL method

First: 3x · 4x = 12x²

Outer: 3x · 1 = 3x

Inner: (-2) · 4x = -8x

Last: (-2) · 1 = -2

12x² + 3x - 8x - 2

Combine like terms

12x² + (3x - 8x) - 2

\(\boxed{12^2-5x-2}\)

Hope this helps!! :)

Please let me know if you have any questions

Answer: 1. 25 percent

2. 1/5

3. 8

4. 0.75

5. about 8 percent

6. 2.4 hrs

7. 62.5 percent

8. Make a porprotion and solve by cross multiplying

9.54$

10. 25

Step-by-step explanation:

Hello!

Answer:

\(\huge\boxed{V = 60 m^{3}}\)

Formula for the volume of a triangular pyramid:

V = 1/3(bh)

The base is a triangle, so b = 1/2(b · h)

Solve for the base:

b = 1/2(8 · 5)

b = 1/2(40)

b = 20 m²

Solve for the volume:

V = 1/3(20 · 9)

V = 1/3(180)

V = 60 m³.

Hope this helped you!

Answer:

60 cubic meters

Step-by-step explanation:

I used the formula V = 1/3 AH to figure it out

All the values are,

a)  lim x → 3 [ 2 f (x) - g (x)] = 18

b)  lim x → 3 [ 2 g (x) ]² = 16

c)  lim x → 3 [ ∛ f (x) / g (x) ] +  lim x → 3 [ 4 h (x) / x + 7 ] = - 1

We have to given that;

Limits are,

lim x → 3 f (x) = 8

lim x → 3 g (x) = - 2

lim x → 3 h (x) = 0

Now, We can simplify all the limits as;

1) lim x → 3 [ 2 f (x) - g (x)]

⇒  lim x → 3 [ 2 f (x)] -  lim x → 3 [ g (x) ]

⇒ 2  lim x → 3 [  f (x) ] - (- 2)

⇒ 2 × 8 + 2

⇒ 16 + 2

⇒ 18

2)  lim x → 3 [ 2 g (x) ]²

⇒ 4 [ lim x → 3  g (x) ]²

⇒ 4 × (- 2)²

⇒ 4 × 4

⇒ 16

3)  lim x → 3 [ ∛ f (x) / g (x) ] +  lim x → 3 [ 4 h (x) / x + 7 ]

⇒ ∛8 / (- 2) + 4 × 0 / (3 + 7)

⇒ - 2/2 + 0

⇒ - 1

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Answer: \(305\)

Shown work is on the image below.

The flow along the curve of intersection between the cylinder and the plane can be found by integrating the dot product of the given velocity field and the curve parameterization. The integral from (0,0,0) to (1,1,1) yields the desired result.

To find the flow along the curve of intersection, we need to parameterize the curve. Let's define = as the parameter, where 0 ≤  ≤ 2π. The curve can be expressed as () = (2cos(), 2sin(), ). Now we can calculate the velocity vector at each point on the curve by substituting the parameterization into the given velocity field. The velocity vector becomes () = (2cos() + 2, -2sin() - 1, 1).

Next, we need to calculate the dot product of the velocity vector and the differential element of the curve parameterization. The differential element is given by  d = (−2sin(), 2cos(), 1). Taking the dot product of () and d, we have () ∙ d = (2cos() + 2)(−2sin()) + (-2sin() - 1)(2cos()) + (1)(1) = -4sin()cos() + 4cos()sin() - 2sin() - 2cos() + 1.

Now we can integrate the dot product over the range of the parameter from 0 to 2π. The integral of () ∙ d with respect to  is equal to the flow along the curve of intersection. Evaluating the integral, we get ∫(()) ∙ d = ∫(-4sin()cos() + 4cos()sin() - 2sin() - 2cos() + 1) d = 2π.

Therefore, the flow along the curve of intersection between the cylinder and the plane from (0,0,0) to (1,1,1) is 2π.

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Answer:

B

Step-by-step explanation:

edge 2021

Answer:

b

Step-by-step explanation:

Answer:

Solution given:

1:

<5=<6

<5+<4=180°[co interior angle]

Substituting value of<5

<6+<4=180°[it shows a property of co interior angle]

So

l || m

2:

<1=90°[ l is perpendicular to t]

<2=90°[m is perpendicular to t]

since

<1=<2[shows property of corresponding angle]

:.

l || m.

3:

<1=<2

<1=<3

substituting value of<1 in second one

<2=<3[which shows property of alternate Angel]

So

Segment ST || segment UV.

4:

<RSP=<PQR......[I]

<QRS+<PQR=180°.....[ii]

from equation I and ii we get

<RSP+<QRS=180°[which shows property of co interior angle ]

So

Segment PS || segment QR

These pictures are the questions given in the pdf, let's get the solutions.

1) Solution

It is given that,

→ <5 = <6

Then the co interior angles,

→ <5+ <4 = 180°

Now substituting value of <5,

→ <6+ <4 = 180°

This shows property of co interior angle.

Therefore, L II m.

2) Solution

Take it as,

→ <1= 90°

In above eq. L is perpendicular to t.

→ <2 = 90°

In above eq. m is perpendicular to t.

Then it will be,

→ <1 = <2

It shows property of corresponding angle.

Therefore, L II m.

3) Solution

It is given that,

→ <1 = <2 and <1 = <3

Now substitute,

The value of <1 in second one,

→ <2 = <3

This shows property of alternate angle.

Therefore, ST II UV.

4) Solution

It is given that,

→ <RSP = <PQR --- (1)

→ <QRS + <PQR = 180° --- (2)

Now from the equation (1) and (2),

→ <RSP + <QRS = 180°

It shows property of co interior angle.

Therefore, PS II QR.

Answer:

161

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

Full equation: y = 3/2x + 1

Answer: if u looking for x it - 1

Answer:

x=-1

Step-by-step explanation:

30=25-5x

5=-5x

-1=x

Answer:

8x=12 cause y would cancel out and then you just add 2 and 6, and 7 and 5.

Answer:

x=1+√3±√4-2√3 / 2

Step-by-step explanation:

hopefully ur able to type this much! happy to answer any questions abt this

If d(n) is O(f(n)) and e(n) is O(g(n)), the product d(n)e(n) is O(f(n)g(n)).

To mathematically show that if d(n) is O(f(n)) and e(n) is O(g(n)), the product d(n)e(n) is O(f(n)g(n)), we can use the definition of big O notation.

Let's assume that d(n) = O(f(n)) and e(n) = O(g(n)). This means that there exist positive constants c1, c2, n0, and n'0 such that for all n ≥ n0 and n' ≥ n'0:

|d(n)| ≤ c1|f(n)| (1)

|e(n)| ≤ c2|g(n)| (2)

We want to show that the product d(n)e(n) is O(f(n)g(n)). To do this, we need to find positive constants c and n'' such that for all n ≥ n'':

|d(n)e(n)| ≤ c|f(n)g(n)|

Now, we can write the product d(n)e(n) as:

|d(n)e(n)| = |d(n)||e(n)|

Using inequalities (1) and (2), we can substitute them into the above expression:

|d(n)e(n)| ≤ c1|f(n)|c2|g(n)|

Let c = c1c2 and n'' = max(n0, n'0). Then for all n ≥ n'':

|d(n)e(n)| ≤ c|f(n)g(n)|

This shows that the product d(n)e(n) is O(f(n)g(n)).

Therefore, we have mathematically shown that if d(n) is O(f(n)) and e(n) is O(g(n)), the product d(n)e(n) is O(f(n)g(n))

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