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3 years ago

I need help on all of these plz

Mathematics

1 answer:

shusha [124]3 years ago

8 0

Can't read the top questions mate !

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The length of a rectangle is three times the width. Ten area of the rectangle is 108 sq in. What is the width of the triangle? A

zubka84 [21]

the answer would be a. 6"

4 0

3 years ago

In the following figure, line DG is tangent to point A at F; line EC and line EH are tangent segments. If CE = 18, find HE.

gogolik [260]

lengths of tangents drawn from an external point are equal ⇒
HE = CE = 18 units.

8 0

3 years ago

What is the Y intercept form if the points go through -8, -1

bogdanovich [222]

the y intercept would be -1

5 0

3 years ago

Solve for z: -21 = z - (-6 - 2z)

Pepsi [2]

Answer:

z = -9

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Algebra I

Terms/Coefficients/Degrees

Step-by-step explanation:

Step 1: Define

-21 = z - (-6 - 2z)

Step 2: Solve for z

Distribute negative: -21 = z + 6 + 2z
Combine like terms: -21 = 3z + 6
Isolate z term: -27 = 3z
Isolate z: -9 = z
Rewrite: z = -9

6 0

3 years ago

Read 2 more answers

If a point P(x,y) is the equidistant from the points A(2,3) and B(6,1), find the equation of locus of moving point P.

jekas [21]

Answer:

$y=2x-6$

Step-by-step explanation:

A locus can be defined as a curve or figure formed by all the points satisfying a particular equation of the relation between coordinates.

The condition stated in the question is such that a generic (x,y) point of the curve is equidistant from the points A(2,3) and B(6,1).

The distance d1 from (x,y) to (2,3) is:

$d_1=\sqrt{(x-2)^2+(y-3)^2}$

The distance d2 from (x,y) to (6,1) is:

$d_2=\sqrt{(x-6)^2+(y-1)^2}$

Since d1=d2:

$\sqrt{(x-2)^2+(y-3)^2}=\sqrt{(x-6)^2+(y-1)^2}$

Squaring both sides:

$(x-2)^2+(y-3)^2=(x-6)^2+(y-1)^2$

Operating:

$x^2-4x+4+y^2-6y+9=x^2-12x+36+y^2-2y+1$

Simplifying all the squares:

$-4x+4x+4-6y+9=-12x+36-2y+1$

Moving the variables to the left side and the numbers to the right side:

$-4x+12x-6y+2y=36+1-4-9$

Simplifying:

$8x-4y=24$

Dividing by 4:

$2x-y=6$

Or, equivalently:

$\boxed{y=2x-6}$

7 0

3 years ago