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1 year ago

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Solve (x + 1)2 – 4(x + 1) + 2 = 0 using substitution. u = x 4(x +1) x +1 (x + 1)2

1 answer:

marta [7]1 year ago

4 0

no substituion only rea numbersStep-by-stion:

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Given that r||s and q is a transversal, we know that by the [________]. corresponding angles theorem alternate interior angles t

neonofarm [45]

Answer:

alternate interior angles theorem

Step-by-step explanation:

The alternate interior angles theorem states that when two parallel lines are cut by a transversal, the resulting angles produced are a pair of congruent alternate interior angles.

Given the image attached below, both line k and line l are parallel to each other and also, line t is the transversal, therefore the resulting congruent alternate interior angles produced are:

∠ 4 ≅ ∠6, ∠1 ≅ ∠7

5 0

2 years ago

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must a function that is decreasing over a given interval always be negative over that same interval explain

kolbaska11 [484]

check the picture below.

notice, a function that is decreasing, namely have a negative slope, doesn't have to be negative.

8 0

2 years ago

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Stores often mark up items each year to cover inflation. Last year, a CD at Juan’s Tunes cost $12.95. This year, the cost was ma

Ratling [72]

4*12.95*1.08-4*12.95*1.08*.10
if u buy 4

7 0

1 year ago

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Three times the sum of half Carlita's age and 3 is at least 12. What values represent Carlita's possible age?

cestrela7 [59]

Answer:

at least 6

Step-by-step explanation:

3(.5x+3)>12

1.5x+9>12

1.5>9

x>6

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1 year ago

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Helen’s house is located on a rectangular lot that is 1 1/8 miles by 9/10 mile.Estimate the distance around the lot

RUDIKE [14]

Answer:

The distance around the lot is 4.05 miles .

Step-by-step explanation:

Formula

$Perimeter\ of\ rectangle = 2 (Length + Breadth)$

As given

$Helen’s\ house\ is\ located\ on\ a\ rectangular\ lot\ that\ is\ 1 \frac{1}{8}\ miles\ by\ \frac{9}{10}\ mile.$

i.e

$Helen’s\ house\ is\ located\ on\ a\ rectangular\ lot\ that\ is\ \frac{9}{8}\ miles\ by\ \frac{9}{10}\ mile.$

Here

$Length = \frac{9}{8}\ miles$

$Breadth = \frac{9}{10}\ miles$

Put in the formula

$Perimeter\ of\ rectangle = 2(\frac{9}{8}+\frac{9}{10})$

L.C.M of (8,10) = 40

$Perimeter\ of\ rectangle = \frac{2\times (9\times 5+9\times 4)}{40}$

$Perimeter\ of\ rectangle = \frac{2\times (45+36)}{40}$

$Perimeter\ of\ rectangle = \frac{2\times 81}{40}$

$Perimeter\ of\ rectangle = \frac{81}{20}$

Perimeter of rectangle = 4.05 miles

Therefore the distance around the lot is 4.05 miles .

6 0

2 years ago

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