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julia-pushkina [17]

2 years ago

5

Rectangle EFGH is translated according to the rule T–5, 9(x, y). If the coordinates of the pre-image of point H are (–2, –3), wh

at are the coordinates of H'?

2 answers:

photoshop1234 [79]2 years ago

7 0

H' = -2 -5, -3 + 9 = (-7, 6)

tatyana61 [14]2 years ago

3 0

Answer:

(-7,6)

Step-by-step explanation:

Just took test on edgen

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The product of a binomial and a trinomial is x 3 + 3 x 2 − x + 2 x 2 + 6 x − 2. Which expression is equivalent to this product a

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Answer:

$x^3+3x^2+5x+2$

Step-by-step explanation:

The product of a binomial and a trinomial is

$x^3+3x^2-x+2 \times 2 +6x-2$

we have to simplify that

$x^3+3x^2-x+4+6x-2$

bringing the like terms together we get

$x^3+3x^2-x+6x+4-2$

$-x+6x = 5x$

$-2+4 = 2$

hence our expression becomes

$x^3+3x^2-x+6x+4-2=x^3+3x^2+5x+2$

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Express the number 220 as the sum of four numbers that form a geometric progression such that the third term is greater than the

siniylev [52]

Answer:

Step-by-step explanation:

The first four terms of geometric series is:

$a,ar,ar^2,ar^3$

Since, we have given information that the third number is greater than 44 that means:

$ar^2=a+44$

Above equation can be rewritten as:

$a(r^2-1)=44$

Now, using:

$a^2-b^2=(a+b)(a-b)$

Here, a=r,b=1

$a(r+1)(r-1)=44$ (1)

The sum of first four terms is:

$a+ar+ar^2+ar^3=220$

$a(1+r+r^2+r^3)=220$

$a(1(1+r)+r^2(1+r))=220$

$a((1+r)(1+r^2))=220$ (2)

Divide equation (2) by (1) we get:

$\frac{r^2+1}{r-1}=\frac{220}{44}$

$r^2+1=5r-5$

$\Rightarrow r^2-5r+6=0$

$\Rightarrow r^2-3r-2r+6=0$

$\Rightarrow r(r-3)-2(r-3)=0$

$\Rightarrow (r-2)(r-3)=0$

$\Rightarrow r=2,3$

CASE1: When r=2 in $ar^2=a+44$

$a(4)=a+44$

$3a=44$

$a=\frac{44}{3}$

CASE2:When r=3 in $ar^2=a+44$

$a(3)^2=a+44$

$\Rightarrow 9a=a+44$

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$\Rightarrow a=\frac{44}{8}=\frac{11}{2}$

The series becomes:

From CASE1: $\frac{44}{3},\frac{44\cdot 2}{3},\frac{44\cdot 2^2}{3},\frac{44\cdot 2^3}{3}$

$\Rightarrow \frac{44}{3},\frac{88}{3},\frac{176}{3},\frac{352}{3}$

From CASE2: $\frac{11}{2},\frac{11\cdot 3}{2},\frac{11\cdot 3^2}{2},\frac{11\cdot 3^3}{2}$

$\Rightarrow \frac{11}{2},\frac{33}{2},\frac{99}{2},\frac{297}{2}$

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