MATHEMATICS HIGH SCHOOL

If ABC ~ DEC, solve for x. The image is not drawn to scale.A.
x = 2
B.
x = 3
C.
x = 4
D.
x = 13

Answers

Answer 1

Answer:

Answer: A. x = 2

Step-by-step explanation:

In the given picture we have $\triangle{ABC}\sim\triangle{DEC}$

Since, we know that the corresponding sides in similar triangles are in proportion.

Therefore, we have

$(CD)/(AC)=(CE)/(BC)\n\n\Rightarrow(8+x)/(6)=(19-2x)/(11-x)\n\n\Rightarrow\ (11-x)(8+x)=6(19-2x)\n\n\Rightarrow\ 88 + 3x -x^2=114-12x\n\n\Rightarrow x^2-15x+26=0\n\n\Rightarrow\ (x-13)(x-2)=0\n\n\Rightarrow\ x=13,2$

But x can not be 13 because BC=11-13=-2, which is not possible.

Therefore, the value of x=2.

Answer 2

Answer: $If\ \Delta ABC\sim\Delta DEC\ then:\n\n(19-2x)/(11-x)=(8+x)/(6)\ where\ x\in(-8;\ 11)\ \ \ |cross\ multiply\n\n(11-x)(8+x)=6(19-2x)\n11(8)+11(x)-x(8)-x(x)=6(19)+6(-2x)\n88+11x-8x-x^2=114-12x\n-x^2+3x+88=114-12x\ \ \ |subtract\ 114\ from\ both\ sides\n-x^2+3x-26=-12x\ \ \ \ \ |add\ 12x\ to\ both\ sides\n-x^2+15x-26=0\ \ \ \ |change\ signs\nx^2-15x+26=0\nx^2-2x-13x+26=0\nx(x-2)-13(x-2)=0\n(x-2)(x-13)=0\iff x-2=0\ or\ x-13=0\n\nx=2\in(-8;\ 11)\ or\ x=13\notin(-8;\ 11)\n\nAnswer:\boxed{\boxed{A.\ x=2}}$

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If x + y = 13 and y = 2, then which of the following is true according to thesubstitution property?
OA) 2 + y = 13
B) x= 13- y
OC) x+ 2 = 13
OD) y = 13 - x

Answers

Answer:

Step-by-step explanation:

Somebody please HELP I NEED TO PASS THIS

Answers

Y is equal to 21 and x is equal to 10

C=25. S=a. B=4c-s2. What would b equal to

Answers

Answer: B = 100 - a^2

Solution:

Step 1: Substitute all the values that are given in the equation

that would be C = 25 and S = a

Step 2: when substituted the equation would be

B = 4 (25) - (a)^2

B = 100 - a^2 Would be the answer

Note: if there was a numerical value to s then the answer would be a number but since s=a so the answer has variable too.

For a cone, if the radius was quadrupled and the slant height was reduced to 1/6 of its original size what would be the formula to find the modified surface area.

Answers

Let $r$ be the radius of the cone's base and $\ell$ the slant height.

The surface area of a cone is

$A=\pi r^2+\pi r\ell$

If the radius is quadrupled ( $r\to4r$ ) and the slant height scaled by a factor of $\frac16$ , then the new area is

$A=\pi(4r)^2+\pi (4r)\left(\frac\ell6\right)$
$A=16\pi r^2+\frac23\pi r\ell$

PLEASE HELP MEEEEEEEEEEEEE

Answers

Answer: 20 ft

Step-by-step explanation:

The mirror is set away from the flagpole on the same scale set away from her.

The distance she stand's away from the mirror is 1/4 of the length of the distance between the mirror and the flag pole because she's 1/4 of the flagpole's size.

This means her size x4 is the size for the flag pole. This checks out because 12x4=48

So to find the height we must do 5x4

5x4=20

A dance floor is made from square wooden tiles with a side length of 1 foot. the floor can be laid out as a square or as a rectangle. Yeasmin chooses a rectangular dance floor for her party. The width of the rectangular dance floor is 10 feet less than the width of the square dance floor, and its length is 15 greater than the length of the square floor.how much greater is the length of the rectangular dance floor than its width?
The perimeter of the rectangular dance floor is 130 feet. What are the length and width of the dance floor?
If the unit rate is $0.50 per tile, how much would it cost Yeasmin to buy the number of tiles needed to make up the dance floor

Answers

legnth/width for square = x
width of rectangle= x-10
legnth of rectangle=x+15
so (x-10)+25=x+15 which is the first part

part two
the perimiter of rectangle floor =130
perimiter= 2width+2legnth so 130/2=width+height
width=x-10 and legnth=x+15 so
x-10+x+15=130/2
2x+5=75
minus 5 from both sides
2x=70
divide by two
x=35 which is legnth/width of the square, but we want the legnth and width of the rectangle so
x-10=width=25
x+15=legnth=50
so the cost of the floor is (50 times 25)times $0.50 or
(50 times 25)/2 so
$525

If ABC ~ DEC, solve for x. The image is not drawn to scale.A.x = 2B.x = 3C.x = 4D.x = 13

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If ABC ~ DEC, solve for x. The image is not drawn to scale.A.
x = 2
B.
x = 3
C.
x = 4
D.
x = 13