MATHEMATICS HIGH SCHOOL

A four-sided shape with the top side labeled as 10.2 cm. The height is labeled 5 cm. A portion of the base from the perpendicular to a vertex is labeled 4 cm. The portion of the base from the perpendicular to the right vertex is 6.2 cm.What is the area of the figure?

25.5 cm2
45.5 cm2
51 cm2
56.1 cm2

Answers

Answer 1

Answer:

The area of the figure is 51 cm².

Given that,

A four-sided figure with the top side labeled as 10.2 cm.

The height is labeled 5 cm.

This shape is a trapezium.

Total length of the base = 6.2 + 4 = 10.2 cm

Area of the figure = 10.2 × 5 = 51 cm²

Hence the required area of the figure is 51 cm².

Learn more about Area here :

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The expression 8 x plus 12 y represents the sum of Harry and Mike’s total monthly wages, where x represents the number of hours Harry worked and y represents the number of hours Mike worked. What is another way to write the expression, and what can you conclude from rewriting it in this way?A. 8 left parenthesis x plus 1.5 y right parenthesis ; Harry’s hourly wage is 1.5 times Mike’s.B. 8 left parenthesis x plus 1.5 y right parenthesis; Mike’s hourly wage is 1.5 times Harry’s.C. 12 left parenthesis 8 x plus y right parenthesis ; Harry’s hourly wage is 8 times Mike’s.D. 8 left parenthesis x plus 4 y right parenthesis ; Mike’s hourly wage is 4 times Harry’s.

The center of a cricle ids (h,7) and the radius is 10. The circle passes through (3,-1). Find all possible values of h.

Answers

$center\ of\ a\ circle:(a;\ b)\n\nradius:r\n\n(x-a)^2+(y-b)^2=r^2\n\n========================\n\n(h;\ 7);\ r=10\n\n(x-h)^2+(y-7)^2=10^2\n\n========================$

$The\ circle\ passes\ throught\ (3;-1).\n\nSubstitute\ x=3\ and\ y=-1:\n\n(3-h)^2+(-1-7)^2=100\n\n(3-h)^2+(-8)^2=100\n\n(3-h)^2+64=100\ \ \ /-64\n\n(3-h)^2=36\iff3-h=\pm√(36)\n\n3-h=-6\ or\ 3-h=6\n-h=-6-3\ or\ -h=6-3\n-h=-9\ or\ -h=3\nh=9\ or\ h=-3\n\nAnswer:h=9\ or\ h=-3$

$equation\ of\ a\ circle:\n\n(x-h)^2+(y-7)^2=10^2\n\nif\ \ (x;y)=(3;-1),\ then\ \ (3-h)^2+(-1-7)^2=100\n\n(3-h)^2+64=100\n\n (3-h)^2=36\ \ \ \Leftrightarrow\ \ \ (3-h=6\ \ \ or\ \ \ 3-h=-6)\n\n3-h=6\ \ \ \ \ \Rightarrow\ \ \ h=-3\n3-h=-6\ \ \ \Rightarrow\ \ \ h=9\n\nAns.\ h=-3\ \ or\ \ h=9$

Someone please help me!!!

Answers

Answer:

thats easy you don't need help

Step-by-step explanation:

Kelly works in the clothing store where she earns a 20% commission on the clothes she sells. Last Saturday she sold $500 in merchandise. what is her commission for Saturday?

Answers

her commission in saturday is 20%*500 / 0.2*500=$100

permutations 2.4 statistics answers A clothing store has a certain shirt in 4 sizes: small, medium, large, and extra large. If it has 2 small, 3 medium, 6 large, and 2 extra large in stock, in how many orders can it sell all the shirts?

Answers

Answer: 360,360

Step-by-step explanation:

The number of permutations of n things of which a are identical, b are identical and so on... :-

$(n!)/(a!\ b!....)$

Given : A clothing store has a certain shirt in 4 sizes: small, medium, large, and extra large.

If it has 2 small, 3 medium, 6 large, and 2 extra large in stock, then total shirts : $2+3+6+2=13$

Then, the number of orders in which the store can sell all the t-shirts will be :_

$(13!)/(2!3!6!2!)\n\n=(13*12*11*10*9*8*7*6!)/(2*3*2*6!*2)\n\n=360360$

Hence, the number of orders in which the store can sell all the t-shirts = 360,360

How many rectangles can you build with a prime number of square tiles?

Answers

You can only make 1 rectangle because 2 square tiles would make a rectangle and 2 is the only prime number that could be used to make a rectangle if the prime number is the amount of tiles used.

Two buildings on opposites sides of a highway are 3x^3- x^2 + 7x +100 feet apart. One building is 2x^2 + 7x feet from the highway. The other building is x^3 + 2x^2 - 18 feet from the highway. What is the standard form of the polynomial representing the width of the highway between the two building?

Answers

Given:

Distance between two buildings = $3x^3- x^2 + 7x +100$ feet apart.

Distance between highway and one building = $2x^2 + 7x$ feet.

Distance between highway and second building = $x^3 + 2x^2 - 18$ feet.

To find:

The standard form of the polynomial representing the width of the highway between the two building.

Solution:

We know that,

Width of the highway = Distance between two buildings - Distance of both buildings from highway.

Using the above formula, we get the polynomial for width (W) of the highway.

$W=3x^3- x^2 + 7x +100-(2x^2 + 7x)-(x^3 + 2x^2 - 18)$

$W=3x^3- x^2 + 7x +100-2x^2-7x-x^3 -2x^2+18$

Combining like terms, we get

$W=(3x^3-x^3)+(- x^2 -2x^2-2x^2)+ (7x -7x)+(100 +18)$

$W=2x^3-5x^2+0+118$

$W=2x^3-5x^2+118$

Therefore, the width point highway is $2x^3-5x^2+118$ .

The correct standard form of the polynomial equation that represents the width of the highway between the two buildings is: $2x^3-5x^2+118$ .

Given:

Distance between the building: $3x^3-x^2+7x+100$