Classify each pair of labeled angles as complementary, supplementary, or neither. Drag and drop the choices into the boxes to correctly complete the table. Each pair of angles may belong to more than one category. complementary supplementary neither Two angles, on the left a fifty three degree angle and on the right a one hundred twenty seven degree angleTwo angles, on the top a thirty four degree angle and on the bottom a fifty six degree angleTwo lines intersect with an angle on the top left labeled sixty seven degrees and on the bottom right sixty seven degrees

Area of a triangle = [tex]\frac{1}{2} base*height[/tex]

In a right angle triangle the two legs are the base and height of the triangle.

One angle of the triangle is 23 degree  and one adjacent leg is 27.6 cm.Let the other leg opposite to 23 degree angle by x.Sine of an angle is ratio of opposite and hypotenuse. Hypotenuse is given as 30 cm.

[tex]Sin 23=\frac{x}{30}[/tex]

x= 30 sin23.

x=11.72cm.

The legs of the triangle are 11.72 cm and 27.6 cm.

Area of triangle =[tex]\frac{1}{2} x11.72 x27.6 =161.7cm^{2}.[/tex]

Sum of two monomials is not necessarily always a monomial.

For example:

Suppose we have two monomials as 2x and 5x.

Adding 2x+5x , we get 7x.

So if two monomials are both like terms then their sum will be a monomial.

Suppose we have two monomials as 3y and 4x

Now these are both monomials but unlike, so we cannot add them together and sum would be 3y + 4x , which is a binomial.

So if we have like terms then the sum is monomial but if we have unlike terms sum is binomial.

Product of monomials:

suppose we have 2x and 5y,

Product : 2x*5y = 10xy ( which is a monomial)

So yes product of two monomials is always a monomial.

The sum of two monomials is not always a monomial and the product of two monomials is always a monomial.

Further explanation:

Explanation:

Consider the two monomials as [tex]5y{\text{ and }}7y.[/tex]

The sum of the two polynomials can be obtained as follows,

[tex]\begin{aligned}{\text{Sum}} &= 5y + 7y\\&= 12y\\\end{aligned}[/tex]

The sum of the two monomials is monomial.

Consider the two monomials as [tex]5y{\text{ and }}7x.[/tex]

The sum of the two polynomials can be obtained as follows,

[tex]{\text{Sum}} = 5y + 7x[/tex]

The sum of the two monomials is not monomial.

Consider the two monomials as [tex]4y{\text{ and }}6x.[/tex]

The product of the two polynomials can be obtained as follows,

[tex]\begin{aligned}{\text{Product}} &= 4y \times 6x\\&= 24xy\\\end{aligned}[/tex]

The product of the two monomials is monomial.

The sum of two monomials is not always a monomial and the product of two monomials is always a monomial.

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Answer details:

Grade: High school

Subject: Mathematics

Chapter: Number system

Keywords: sum, two monomial, monomials, always monomial, product, always, monomial, factor, factorization, polynomial, quadratic, cubic, greatest common factor, groups, multiplication, product, identities, common factor, expression, terms, grouping.

Answer:

the answer is V≈19.68cm³

V=a3

a

Edge

cm

Step-by-step explanation:

Final answer:

To find out how much more was cut from the skirt than the pants, convert fractions to have a common denominator and subtract: 3/4 inch minus 1/6 inch equals 7/12 inch.

Explanation:

To determine how much more the tailor cut off the skirt than the pants, we need to subtract the length cut from the pants from the length cut from the skirt. The tailor cut 3/4 of an inch off the skirt and 1/6 of an inch off the pants. To perform the subtraction, we need a common denominator, in this case, 12 is suitable. So we convert 3/4 to 9/12 and 1/6 to 2/12.

Now, we subtract the smaller fraction from the larger one:

9/12 - 2/12 = 7/12 of an inch.

Therefore, the tailor cut 7/12 of an inch more off the skirt than the pants.

The equation to represent the relationship between the time Sabar walks, denoted by [tex]x[/tex] , and the number of calories he burns, denoted by [tex]y[/tex], can be written as [tex]\( y = mx \),[/tex] where [tex]\( m \)[/tex] is the rate at which calories are burned per unit time.

To establish a relationship between the time Sabar spends walking and the number of calories he burns, we can use a simple linear equation. In this context, the independent variable [tex]x[/tex] represents the time spent walking, and the dependent variable [tex]\( y \)[/tex] represents the number of calories burned.

The rate at which calories are burned is typically constant for a given activity if the intensity remains consistent. This rate is represented by the coefficient [tex]\( m \)[/tex] in the equation [tex]\( y = mx \)[/tex]. The value of [tex]\( m \)[/tex] would be determined by how many calories Sabar burns per minute, hour, or any other time unit chosen for [tex]x.[/tex]

For example, if Sabar burns 5 calories per minute, then [tex]\( m = 5 \)[/tex]calories/minute, and the equation would be [tex]\( y = 5x \)[/tex]. If we measure time in hours and Sabar burns 300 calories per hour, then [tex]\( m = 300 \)[/tex]calories/hour, and the equation would be [tex]\( y = 300x \).[/tex]

This linear equation assumes that the relationship between time and calories burned is directly proportional, meaning that the number of calories burned increases at a constant rate as the time spent walking increases. This is a common assumption for steady-state aerobic activities like walking at a consistent pace.

The row of 9 bolts laid end to end would be 7 7/8 inches long.

To find the total length of 9 bolts laid end to end, we multiply the length of one bolt by the number of bolts.

Given:

- Length of one bolt: [tex]\( \frac{7}{8} \)[/tex] inch

- Number of bolts: 9

Total length:

[tex]\[ \text{Total length} = \text{Length of one bolt} \times \text{Number of bolts} \][/tex]

[tex]\[ \text{Total length} = \frac{7}{8} \times 9 \][/tex]

Now, multiply:

[tex]\[ \text{Total length} = \frac{7 \times 9}{8} \][/tex]

[tex]\[ \text{Total length} = \frac{63}{8} \][/tex]

To express this in mixed number form:

[tex]\[ \frac{63}{8} = 7 \frac{7}{8} \][/tex]

Therefore, the row of 9 bolts laid end to end would be [tex]\( 7 \frac{7}{8} \)[/tex] inches long.

Answer:

[tex]Arc length=32.97m[/tex]        

Step-by-step explanation:

It is given that In a circle with a radius of 6 m, an arc is intercepted by a central angle of [tex]\frac{7\pi}{4}[/tex] radians.

Then, the arclength of the circle is given as:

[tex]Arc length=radius{\times}central angle[/tex]

⇒[tex]Arc length=6{\times}\frac{7\pi}{4}[/tex]

⇒[tex]Arc length=\frac{21\pi}{2}[/tex]

⇒[tex]Arc length=10.5{\times}3.14[/tex]

⇒[tex]Arc length=32.97m[/tex]

Thus, the arc length of the given circle is 32.97m.

Given is the damage index equation, [tex] y=\frac{-2}{3} x+14 [/tex]

where y represents damage index, and x represents ambient temperature.

It is similar to slope-intercept form of linear equation given by y = mx + b.

Where m is said to be the slope of line and b is the y-intercept of the line. When x = 0, y = b.

Similarly, if plug x = 0 in the damage index equation, [tex] y=\frac{-2}{3} (0)+14=14 [/tex]

So, If ambient temperature, x = 0. Then damage index, y = 14.

Hence, option D is correct i.e. " If the ambient temperature is 0 then the damage index would be 14".

Answer:

The answer is D because 14 is the y-intercepts, which means that the x-value is 0. So the correct meaning is If the ambient temperature is 0 then the damage index would be 14. .

Step-by-step explanation: