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2 years ago

Quadrilateral CDEF is reflected across the x-axis and translated 3 units left to create quadrilateral GHJK .

Mathematics

1 answer:

PilotLPTM [1.2K]2 years ago

6 0

Answer- KG⎯∥JH⎯

the given figure is a trapezium in which CF||DE

Reflection of figure CDEF and shifting it by 3 units does not change the shape of the figure but changes only its position in the x-y plane.

so, the shape of the figure GHJK will be same as that of CDEF.

so sides, KG⎯∥JH⎯

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Answer:

D.) SAS

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2 years ago

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Grade 10 Math. Solve for y. Will mark right answer brainliest :)

Dima020 [189]

Answer:

y=5, y= $\frac{38}{11}$

Step-by-step explanation:

Hi there!

We are given the equation

$\frac{y+2}{y-3}$ + $\frac{y-1}{y-4}$ = $\frac{15}{2}$ and we need to solve for y

first, we need to find the domain, which is which is the set of values that y CANNOT be, as the denominator of the fractions cannot be 0

which means that y-3≠0, or y≠3, and y-4≠0, or y≠4

$\frac{y+2}{y-3}$ and $\frac{y-1}{y-4}$ are algebraic fractions, meaning that they are fractions (notice the fraction bar), but BOTH the numerator and denominator have algebraic expressions

Nonetheless, they are still fractions, and we need to add them.

To add fractions, we need to find a common denominator

One of the easiest ways to find a common denominator is to multiply the denominators of the fractions together

Let's do that here;

on $\frac{y+2}{y-3}$ , multiply the numerator and denominator by y-4

$\frac{(y+2)(y-4)}{(y-3)(y-4)}$ ; simplify by multiplying the binomials together using FOIL to get:

$\frac{y^{2}-2y-8}{y^{2}-7y+12}$

Now on $\frac{y-1}{y-4}$ , multiply the numerator and denominator by y-3

$\frac{(y-1)(y-3)}{(y-4)(y-3)}$ ; simplify by multiplying the binomials together using FOIL to get:

$\frac{y^{2}-4y+3}{y^{2}-7y+12}$

now add $\frac{y^{2}-2y-8}{y^{2}-7y+12}$ and $\frac{y^{2}-4y+3}{y^{2}-7y+12}$ together

Remember: since they have the same denominator, we add the numerators together

$\frac{y^{2}-2y-8+y^{2}-4y+3}{y^{2}-7y+12}$

simplify by combining like terms

the result is:

$\frac{2y^{2}-6y-5}{y^{2}-7y+12}$

remember, that's set equal to $\frac{15}{2}$

here is our equation now:

$\frac{2y^{2}-6y-5}{y^{2}-7y+12}$ = $\frac{15}{2}$

it is a proportion, so you may cross multiply

2(2y²-6y-5)=15(y²-7y+12)

do the distributive property

4y²-12y-10=15y²-105y+180

subtract 4y² from both sides

-12y-10=11y²-105y+180

add 12 y to both sides

-10=11y²-93y+180

add 10 to both sides

11y²-93y+190=0

now we have a quadratic equation

Let's solve this using the quadratic formula

Recall that the quadratic formula is y=(-b±√(b²-4ac))/2a, where a, b, and c are the coefficients of the numbers in a quadratic equation

in this case,

a=11

b=-93

c=190

substitute into the formula

y=(93±√(8649-4(11*190))/2*11

simplify the part under the radical

y=(93±√289)/22

take the square root of 289

y=(93±17)/22

split into 2 separate equations:

y= $\frac{93+17}{22}$

y= $\frac{110}{22}$

y=5

and:

y= $\frac{93-17}{22}$

y= $\frac{76}{22}$

y= $\frac{38}{11}$

Both numbers work in this case (remember: the domain is y≠3, y≠4)

So the answer is:

y=5, y= $\frac{38}{11}$

Hope this helps! :)

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2 years ago

What is the name for the total area of the surface of a solid?prism area net area surface area facial area

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