Help will be appreciated !

Answer:

hi the anser might be 600$

Step-by-step explanation:

Answer:

Graph #1 - y=-20x+40

Graph #2 - y=5x+10

Step-by-step explanation: The first thing that you do is pull two points from each of the graphs and find the slope. To find the slope you use

FROM GRAPH #1:

(x1,y1)    (x2,y2)

   80-40/2-3

      20/-1

    M = -20

FROM GRAPH #2:

(x1,y1)  (x2,y2)

  20-10/3-1

     10/2

    M = 5

2. Now once you find the slope you will use the equation of a line formula to form an equation.

The (mx) is your slope, the (b) is the y-intercept. (any value on the y-axis)

3. Since we now know the equation of a line and the slope we will substitute the variables.

Graph #1:

y=mx+b

y=-20x+40

Graph #2:

y=mx+b

y=5x+10

Answer:

  x = {-2, 0, 2}

Step-by-step explanation:

You want the solutions to √((√(x²) -1)²) = 1.

Squaring both sides gives ...

  (√(x²) -1)² = 1² = 1

Taking the square root of both sides gives ...

  √(x²) -1 = ±1

Adding 1, we have ...

  √(x²) = 1 ± 1 = {0, 2}

We know that √(x²) = |x|, so this reduces to ...

  |x| = {0, 2}

  x = {±0, ±2}

  x = {-2, 0, 2}

__

Additional comment

For finding the solutions graphically, we like to plot the equation in the form f(x) = 0. The graphing calculator readily identifies the x-intercepts, which are the solutions.

Answer:

[tex]\triangle ABC \cong \triangle DCB[/tex] by ASA

Step-by-step explanation:

[tex]\overline{BC} \cong \overline{BC}[/tex], so [tex]\triangle ABC \cong \triangle DCB[/tex] by ASA.

Answer:

a

Step-by-step explanation:

86 and 87 are similar. Also none of the other answers are correct

9514 1404 393

Answer:

  A.  x = {nπ, π/4 +nπ}

Step-by-step explanation:

The zero product rule tells you the solutions are found where the factors are zero:

  tan(x) = 0   ⇒   x = nπ

  tan(x) -1 = 0   ⇒   x = arctan(1) +nπ = π/4 +nπ

_____

When working with functions of trig functions, it is helpful to have an idea of the general shape of their graphs and a few "special" values.

The value of the variable x and y will be 1.5 and -2.5, respectively. Then the value of the expression (x + y) will be negative 1.

The allocation of weights to the important variables that produce the calculation's optimum is referred to as a direct consequence.

A system of equations is given below.

x = -2y - 3.5       ...1

-5x + 3y = -15    ...2

From equations 1 and 2, then we have

-5(-2y - 3.5) + 3y = -15

10y + 17.5 + 3y = - 15

13y = -32.5

y = - 2.5

Then the value of the variable 'x' is calculated as,

x = - 2(-2.5) - 3.5

x = 5 - 3.5

x = 1.5

Then the value of the expression (x + y) is calculated as,

x + y = 1.5 + (-2.5)

x + y = 1.5 - 2.5

x + y = - 1

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Answer:

Capacity = 9.63509

Step-by-step explanation:

V = πr²h

9.5 x 9.5 x 3.14 = 283.385

283.385 x 34 = 9,635.09

9,635.09 ÷ 1,000 = 9.63509

Please let me know if this helps.

The correct answer is System 1: Infinitely many solutionsSystem 2: One solutionSystem 3: No solution

Let's analyze each system of linear equations and determine the number of solutions they have.

y = x + 6 and 3x - 3y = -18

In this system, the first equation is a linear equation in slope-intercept form, and the second equation is a linear equation in standard form. The two equations represent the same line. Since they are the same line, they intersect at infinitely many points. Therefore, this system has infinitely many solutions.

y = -4x + 11 and -6x + y = 11

Both equations in this system are in slope-intercept form. The slopes of the two lines are different (-4 and -6), indicating that the lines are not parallel. Since the lines are not parallel and have different slopes, they intersect at a single point. Therefore, this system has one solution.

y = -2x + 5 and 2x + y = -7

Both equations in this system are in slope-intercept form. The slopes of the two lines are equal (-2 and 2), but their y-intercepts are different. When the slopes are equal and the y-intercepts are different, the lines are parallel and do not intersect. Therefore, this system has no solution.

Matching the systems with the correct number of solutions:

System 1: Infinitely many solutions

System 2: One solution

System 3: No solution

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Step-by-step explanation:

⭐ What does it mean for a number (x) to be a multiple of another number (y)?

Thus, we have to show that [tex](2n-5)^2 - 13[/tex] is divisible by 4.

Let's compute [tex](2n-5)^2 - 13[/tex]:

[tex](2n-5) ^2 = 4n^2 -20n + 25[/tex]

[tex]4n^2 - 20n + 25 -13[/tex]

[tex]4n^2 - 20n + 12[/tex]

[tex]4n^2 - 20n + 12[/tex] is the expression we have.

To see if this expression is divisible by 4, we need to divide [tex]4n^2 - 20n + 12[/tex] by 4 and have a remainder of 0.

The given equation (2n-5)² - 13 equals 4² - 20n + 12. Because each term in the simplified equation can be divided evenly by 4, it proves that the equation is indeed a multiple of 4, assuming n is an integer.

The problem is asking to show that the expression (2n-5)² - 13 is a multiple of 4 for all integers. To prove this, we need to expand and simplify the expression and then check whether it can be divided evenly by 4.

Let's start by expanding the expression:

(2n-5)² - 13 = 4n² - 20n + 25 - 13 = 4n² - 20n + 12

Now we can observe that all terms in the equation are divisible by 4:

4n²/4 = n², -20n/4 = -5n, and 12/4 = 3

Because all terms in the equation can be divided evenly by 4, we can say that the entire equation (2n-5)² - 13 is a multiple of 4, assuming n is an integer.

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Answer:

The quen is as following:

ABC is a right triangle at C,

Acute angles are in the ratio 5:1, i.e. ∠BAC : ∠ABC = 5:1

If CH is an altitude to AB and CL is an angle bisector of ∠ACB, find m∠HCL.

The solution is:  m∠HCL = 30°

Step-by-step explanation:

See the attached figure.

∵The triangle is right at C   ∴∠C = 90°

∴∠A + ∠B = 90° ⇒(1)

∵ Acute angles are in the ratio 5:1, i.e. ∠BAC : ∠ABC = 5:1

∴∠A = 5 times ∠B

Substitute at (1)

∴ 5 ∠B + ∠B = 90° ⇒⇒⇒ ∴∠B = 15°  and ∠A = 75°

∵CL is an angle bisector of ∠ACB

∴ ∠ACL = 90°/2 = 45°

∵ CH is an altitude to AB ⇒ ∠CHA = 90°

At the triangle AHC:

∠ACH = 180° - (∠CHA + ∠CAH) = 180° - (90° + 75°) = 15°

∴ ∠HCL = ∠ACL - ∠ACH = 45° - 15° = 30°

Answer:

D = -15g² - 20g

-5g × 3g -5g×4

= -15g² -20g

# to put in ² when typing hold 2 for 5seconds...

#like please

The measures in this problem are given as follows:

1. x = 14.

2. m < 1 = 58º.

3. m < 2 = 122º.

Angles 1 and 2 are supplementary, as the angle 1 and the consecutive interior angle with angle 2 are corresponding angles, meaning that they are congruent, and then the angles are supplementary with their consecutive interior angles.

When two angles are supplementary, the sum of their measures is of 180º, hence the measure of x is obtained as follows:

m < 1 + m < 2 = 180

2x + 30 + 8x + 10 = 180

10x + 40 = 180

10x = 140

x = 140/10

x = 14.

Then the angle measures are obtained as follows:

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Answer:

Step-by-step explanation:

First, let us substitute x and y values: -7 = -2(-5) + 4

Next, let us simplify using the substitution property of equality: 14 = -7 (SEE BELOW MORE INFO).

Now, this does not make sense yet because 14 cannot possibly equal -7. Therefore, we must add a b value, therefore leading us to the equation:

14 + b = -7

by simplifying, we can conclude that b = -21

Finally, we can plug in the b-value into our original equation:

y = -2x + 4 - 21

After simplifying, we get y = -2x - 17. When this is graphed, we can see that -2x - 17 intersects (-5, -7).

 The answer is -2x² + 8x + 54 = 0.

This refers to a method used to easily solve the task of repeated addition of the same number. It is used when we need to combine groups of equal sizes.

Given in the question:

[tex]\frac{2x +3 }{x^{2}-x}[/tex] *[tex]\frac{4x^{2 } +8x }{10x +30}[/tex]

Multiply the numerators and denominators and equate the two:

(2x + 3) * (4x² + 8x)= ( x² - x) * (10x + 30)

expanding further we have:

(2x + 3) * 4x² + (2x + 3) *8x = ( x² - x) * 10x + ( x² - x) *  30

8x³+12x² +16x² +24x= 10x³ -10x² + 30x² -30x

Resolving the left-hand side of the equation and the right-hand side and equating to zero we have:

8x³- 10x³ + 12x² +16x² +10x² -30x² +24x +30x

2x³ + 8x² + 54x = 0.

Divide through by x, we have:

[tex]\frac{-2x^{3} }{x}[/tex]  + [tex]\frac{8x^{2} }{x}[/tex] +    [tex]\frac{54x}{x}[/tex] = 0

-2x² + 8x + 54 = 0.

Hence the answer is-2x² + 8x + 54 = 0.

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The linear model that could be used to calculate cost of mailing a large envelope is L(x) = 0.88 + 0.2(x – 1) and used to calculate cost of mailing a package is P(x) = 1.71 + 0.17(x – 3). The weight at which the cost a large envelope and a package will cost the same amount to mail is 17 ounces.

Linear models are used to describe a continuous response variable as a function of one or more predictor variables. Let x be the weight in ounces and L be the cost of mailing a large envelope. Based on the provided information, the cost of mailing a large envelope increases with increasing weight. The cost of mailing one ounce is $0.88 and increased by $0.2 for each additional weight. Hence,

L(x) = 0.88 + 0.2(x – 1)

Let x be the weight in ounces and P be the cost of mailing a package. Based on the provided information, the cost of mailing a package remains the same for weight of 1 to 3 ounces at the cost of $1.71, and then increases by $0.17 for each additional weight. Hence,

P(x) = 1.71 + 0.17(x – 3)

The weight at which a large envelope and a package will cost the same amount to mail is determined by equating the two models.

L(x) = P(x)

0.88 + 0.2(x – 1) = 1.71 + 0.17(x – 3)

0.88 + 0.2x – 0.2 = 1.71 + 0.17x – 0.51

0.68 + 0.2 x = 0.17 x + 1.2

0.2x – 0.17x = 1.2 – 0.68

0.03x = 0.52

x = 17.33 ≈ 17 ounces

Note: The question is incomplete. The complete question probably is: Provided table indicates the U.S. Postal Rates for large envelopes and packages of different weights. a) Create a linear model that could be used to calculate the cost to mail a large envelope, given its weight in ounces. Use L to represent the cost to mail a large envelope and x to represent the envelope weight in ounces. Create a linear model that could be used to calculate the cost to mail a package, given its weight in ounces. Use P to represent the cost to mail a package and x to represent the package weight in ounces. Find the weight at which a large envelope and a package will cost the same amount to mail, assuming your linear models still apply to higher weights.

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Answer: -28

Step-by-step explanation:

Mutiply by 7 on both sides, it cancels out to the left and then -4 (7) equals -28 so n= -28

Answer:

-28

Step-by-step explanation:

Multiply both sides by 7

That cancel outs the denominator 7 out and multiples -4 to -28

now all that's left is 1n or just n = -28

The solution of the system of equations are x = -4 and y=1.

How to solve system of equations?

5x+y=-19  -- (i)

5x-4y=-24 ---(ii)

Subtract the above two equations:

5y = 5

y = 1

Substitute y=1 in equation (ii)

5x - 4(1) = -24

5x = -20

x = -4

Hence, the solution of the system of equations are x = -4 and y=1.

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Answer:

Okay so for this one use the Pythagorean theorem

Step-by-step explanation:

A^2 + 28^2 = 128^2

You are trying to find A.

So 28^2 is 784

128^2 is 16384

Now you will subtract 784 from both sides of the equation your new equation should be this: a^2 = 15600

Now you find the square root of both sides which gives you: A= 124.89 round it to 125

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Answer:

3

Step-by-step explanation:

23

r packages include sample datasets. they also include reusable r functions and documentation about how to use the functions this statement is true.

Function in math is a mathematical relationship between two variables where one is dependent on the other. This relationship can be expressed as an equation, graph, or table. Every input has one and only one output. This means that the output of a function is determined by the input given. Functions can be used to model real-world phenomena, helping us to make predictions and solve problems.
Packages offer a great way to save time and energy when working with data in r.
R packages are a great way to save time and energy when working with data in R. They provide a convenient way to access multiple datasets, functions, and documentation in one place. Additionally, many packages also contain sample data and code examples to help users quickly get up and running with their data analysis.
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