Explain why all linear angle pairs must be supplementary, but all supplementary angles do not have to be linear pairs?

Answer:

Option (a) is correct.

a(x)b(x) = (ab)(x)

To produce a quadratic equation the two linear equation must be multiply.

Step-by-step explanation:

Given :  a(x) and b(x) are linear functions with one variable.

We have to choose from the given option the correct expression that will produce a quadratic equation.

Quadratic equation is an equation which is in the form of [tex]ax^2+bx+c[/tex] where [tex]a\neq0[/tex]

Since given a(x) and b(x)  are linear equation in one variables so to get square term we must multiply the two linear equations.

Let a(x)=3x+1

and b(x)= 8x+1

then when we multiply we get,

[tex]a(x)\times b(x)=(3x+1)\times (8x+1)=24x^2+11x+1[/tex]

Thus, To produce a quadratic equation the two linear equation must be multiply.

Thus, a(x)b(x) = (ab)(x) is correct.

Option (a) is correct.

Answer:

False.

Step-by-step explanation:

Given,

Chord AB Is 3 inches from the center and chord CD is 5 inches from the center of the same circle,

Since, the distance of a chord is inversely proportional to the length of the chord,

That is, the chord that has the smallest distance from the center is largest.

3 < 5

⇒ The distance of chord AB from center < The distance of Chord CD from center

By the above statement,

⇒ The length of chord AB > The length of chord CD.

Hence, the given statement is FALSE.

Answer: B. intersecting chords

Step-by-step explanation:

When two chords intersect then inscribed angles are formed.

We know that the inscribed angle theorem tells that the measure of an inscribed angle is half the measure of its intercepted arc.

The measure of an angle formed by intersecting chords is half the sum of the measures of the intercepted arcs.

The solution of the equation [tex]13+\frac{w}{7}=-18[/tex] is [tex]\boxed{w=-217}[/tex].

Further explanation:

Given:

The equation is [tex]13+\frac{w}{7}=-18[/tex].

Calculation:

Method (1)

The given equation is as follows:

[tex]\boxed{13+\dfrac{w}{7}=-18}[/tex]  

The above equation is a linear equation that has one degree.

The equation with one variable can be solved by moving all terms in to the one side and simplify the equation for the value of variable.

Subtract [tex]13[/tex] on both sides in the equation (1) to obtain the value of [tex]w[/tex] as follows,

[tex]\begin{aligned}13+\dfrac{w}{7}-13&=-18-13\\13-13+\dfrac{w}{7}&=-31\\ \dfrac{w}{7}&=-31\end{aligned}[/tex]  

Now, multiply [tex]7[/tex] on both sides of the above equation as,

[tex]\begin{aligned}7\times \dfrac{w}{7}&=-31\times7\\w&=-217\end{aligned}[/tex]

Therefore, the value of [tex]w[/tex] is [tex]-217[/tex].

Method (2)

To obtain the solution of the equation (1), take least common multiple of the denominator of the left hand side of the equation as,

[tex]\begin{aligned}13+\dfrac{w}{7}&=-18\\ \dfrac{(13\times7)+w}{7}&=-18\\ \dfrac{91+w}{7}&=-18\end{aligned}[/tex]  

Now, multiply by [tex]7[/tex] on both sides of the above equation to obtain the value of [tex]w[/tex] as follows,

[tex]\begin{aligned}7\times\left(\dfrac{91+w}{7}\right)&=7\times(-18)\\91+w&=-126\end{aligned}[/tex]

Subtract [tex]91[/tex] on both sides of the above equation as follows,

[tex]\begin{aligned}91+w-91&=-126-91\\w&=-217\end{aligned}[/tex]

Therefore, the value of [tex]w[/tex] is [tex]-217[/tex].

Thus,  the solution of the equation [tex]13+\dfrac{w}{7}=-18[/tex] is [tex]\boxed{w=-217}[/tex].

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Answer details:

Grade: Middle school

Subject: Mathematics

Chapter: Linear equations

Keywords: Equation, 13+(w/7)=-18, variables, subtract, linear equations, one degree, multiply least common multiple, linear equation, mathematics.

Answer:

>

Step-by-step explanation:

Given : [tex]55?35[/tex]

To Find: Which of the symbols correctly relates the two numbers below?

Solution:

We are given that [tex]55?35[/tex]

Since we know that 55 is greater than 35 .

So, sign of greater than (>)should replace ?

So, [tex]55>35[/tex]

Hence Option F is correct.

So, [tex]55?35[/tex] = [tex]55>35[/tex]

Option F : >

To write an improper fraction as a mixed number, divide the denominator into the numerator.

Image is provided.

Therefore, the improper fraction 11/9 can be written as the mixed number 1 and 2/9.

The range of the function h(x) = -8x is all real numbers, as there are no restrictions on the output the function can produce. This is typical of a linear function.

The range of a function refers to the possible outputs it can produce. In the case of the function h(x) = -8x, it is a linear function, and there are no restrictions on the values that x can take, meaning x can be any real number between -∞ and +∞. The output, h(x), therefore, can be any real number, also between -∞ and +∞, given that it will be the result of -8 multiplied by x. Therefore, the range of the function h(x) = -8x is all real numbers from -∞ to +∞.

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The addition is:

      [tex]=\dfrac{513}{16}\ in.[/tex] or   [tex]32\frac{1}{16}\ in.[/tex]

We are asked to add the expression :

[tex]21\frac{1}{2}\ in.[/tex] and [tex]10\frac{9}{16}\ in.[/tex]

both the numbers are in mixed fraction .

We will first change both the numbers into simple fraction as follows:

[tex]21\frac{1}{2}=\dfrac{21\times 2+1}{2}\\\\\\21\frac{1}{2}=\dfrac{43}{2}\ in.[/tex]

and

[tex]10\frac{9}{16}=\dfrac{10\times 16+9}{16}\\\\10\frac{9}{16}=\dfrac{169}{16}\ in.[/tex]

Hence, the addition of the two numbers is calculated as follows:

[tex]\dfrac{43}{2}+\dfrac{169}{16}\\\\\\=\dfrac{43\times 8+169}{16}[/tex]

( Since, on taking lcm of 2 and 16)

Hence, we get:

[tex]=\dfrac{513}{16}\ in.[/tex]

which in mixed fraction is:

          [tex]32\frac{1}{16}\ in.[/tex]

To solve the inequality -5r + 6 ≤ -5(r+2), the answer is that this inequality has no solution.

Solve the Inequality:

-5r + 6 ≤ -5(r+2)

Distribute the -5 on the right side: -5r + 6 ≤ -5r - 10

Combine like terms: 6 ≤ -10

Since 6 is not less than or equal to -10, the inequality has no solution.

Answer:

[tex]2 \frac{2}{5}[/tex]

Step-by-step explanation:

Try dividing 12 and 5. What a surprise it doesn't work. 5×2=10. 12÷5=10 r2. r2=[tex]\frac{2}{5}[/tex]. 2+[tex]\frac{2}{5}[/tex]=[tex]2 \frac{2}{5}[/tex].

Answer:

[tex]x=\frac{3}{16}[/tex]

Step-by-step explanation:

Given : Exponential function [tex]9^{8x}=27[/tex]

We have to solve the given exponential equation.

Consider the given exponential function [tex]9^{8x}=27[/tex]

[tex]\mathrm{Convert\:}9^{8x}\mathrm{\:to\:base\:}3[/tex]

[tex]9^{8x}=\left(3^2\right)^{8x}[/tex]

Function becomes,

[tex]\left(3^2\right)^{8x}=27[/tex]

Convert 27 to base 3, we have,

[tex]\left(3^2\right)^{8x}=3^3[/tex]

Apply exponent rule, [tex]\left(a^b\right)^c=a^{bc}[/tex]

We have, [tex]3^{2\cdot \:8x}=3^3[/tex]

[tex]\mathrm{If\:}a^{f\left(x\right)}=a^{g\left(x\right)}\mathrm{,\:then\:}f\left(x\right)=g\left(x\right)[/tex]

We have,

[tex]2\cdot \:8x=3[/tex]

Simplify, we have,

[tex]16x=3[/tex]

Thus, [tex]x=\frac{3}{16}[/tex]

Final answer:

The cosine, cosecant, secant, and cotangent functions are all trigonometric functions. They are continuous everywhere except where the denominator is zero.

Explanation:

The cosine, cosecant, secant, and cotangent functions are all trigonometric functions. These functions are defined for all real numbers except when the denominator becomes zero. Therefore, they are continuous everywhere except where the denominator is zero.

For example, the cosine function, denoted as cos(x), is continuous for all real numbers. It has a periodic nature and oscillates between -1 and 1.

Similarly, the cosecant function, denoted as csc(x), is continuous everywhere except at the points where sin(x) is equal to zero. The secant function, denoted as sec(x), is continuous everywhere except at the points where cos(x) is equal to zero. The cotangent function, denoted as cot(x), is continuous everywhere except at the points where tan(x) is equal to zero.

Answer:  The required ratio is 3 : 1.

Step-by-step explanation:  Given that Juanita is 15 years old and her brother is 5 years old.

We are to find the ratio of Juanita's age to her brother's age.

Let x and y represents the ages of Juanita and her brother respectively in years.

Then, according to the given information, we have

[tex]x=15,\\\\y=5.[/tex]

Therefore, the ratio of Juanita's age to her brother's age is given by

[tex]x:y\\\\=\dfrac{x}{y}\\\\\\=\dfrac{15}{5}\\\\\\=\dfrac{3}{1}\\\\=3:1.[/tex]

Thus, the required ratio is 3 : 1.

1/30 in decimal form can be rounded to 0.033.

To convert the fraction 1/30 into decimal form, you simply divide the numerator (1) by the denominator (30). This can be done using a calculator or by manual division.

1 ÷ 30 = 0.03333... (repeating)

So, 1/30 as a decimal is approximately 0.033 when rounded to three decimal places.

Here are the steps to understand it better:

The annual loss of income due to missing one work day every two weeks at $9.29 per hour, working 8 hours per missed day, is $1,932.32.

To calculate the amount of pay being lost each year due to missing one work day every two weeks, we need to consider the weekly work schedule and hourly wage.

The annual loss of income due to missing one day of work every two weeks at an hourly wage of $9.29 is $1,932.32.

The first thing we are going to do for this case is define variables.

We have then:

w: width

l: length

The perimeter is given by:

[tex] 2w + 2l = 520
[/tex]

The area is given by:

[tex] A = w * l
[/tex]

The area as a function of a variable is:

[tex] A (w) = w * (260-w)
[/tex]

Rewriting we have:

[tex] A (w) = -w ^ 2 + 260w
[/tex]

To obtain the maximum area, we derive:

[tex] A '(w) = -2w + 260
[/tex]

We equal zero and clear the value of w:

[tex] -2w + 260 = 0

2w = 260
[/tex]

[tex] w = \frac{260}{2}

w = 130
[/tex]

Then, the length is given by:

[tex] l = 260 - w

l = 260 - 130

l = 130
[/tex]

Finally, the maximum area obtained is:

[tex] A = w * l

A = 130 * 130

A = 16900
[/tex]

Answer:

A retangle that maximizes the enclosed area has a length of 130 yards and a width of 130 yards.

The maxium area is 16900 square yards