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3 years ago

X^(2)+x-20 splitting the middle term

1 answer:

5 0

Are you trying to factorise this?

Here is a way you may not have been taught....
(x + a)(x + b) = x^2 + (a + b)x + ab

So you need to find two factors of -20 that add up to +1
-4 x 5 = -20 and -4 + 5 = 1. Perfect

x^2 + x - 20 = (x - 4)(x + 5)

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Which of the numbers 19 21 23 and 25 has the most factors

belka [17]

It would either be 21 or 25 as they both have the same amount of factors :)

5 0

4 years ago

What is the description of the translation represented by the translation rule (x,y)-(x-2,y-2)

WINSTONCH [101]

Answer:

B. 2 units to the right and 2 units up

Step-by-step explanation:

7 0

3 years ago

Find the point on the line 6x+y=9 that is closest to the point (-3,1).

Snowcat [4.5K]

Basically, you have to find the line through (-3, 1) that is perpendicular to the line 6x + y = 9 and then find where the two lines intersect. Dropping a perpendicular from the point to the line is the shortest distance from the point to the line.

6x + y = 9
y = -6x + 9
slope is -6

perpendicular would have slope 1/6
(y - 1) = (1/6) (x + 3)
y - 1 = (1/6) x + (1/2)
multiplying both sides by 6
6y - 6 = x + 3
x - 6y = -9

Where do these two lines intersect?

From above we know given equation is y = -6x + 9
Plugging into other equation
x - 6(-6x + 9) = -9
x + 36x - 54 = -9
37 x = 45
x = 45/37
y = -6(45/37) + 9
y = -270/37 + (333/37)
y = 63/37

point is (45/37, 63/37)

7 0

3 years ago

Find the magnitude of vector v that has an initial point of (1, 3) and a terminal point of (7, 1)

solmaris [256]

Answer:

It's totally correct with no doubt

3 0

3 years ago

I NEED HELP WILL GIVE BRAINLIEST!! PLEASE HELP ME ASAP!!!!

GenaCL600 [577]

Answer:

A. The length of the second leg is 8.5 inches

B. The length of the three-dimensional diagonal is 9.9 inches

Step-by-step explanation:

Let us revise the relation between the hypotenuse and the two legs of a right triangle

(hypotenuse)² = (vertical leg)² + (horizontal leg)²

∵ The length of the rectangular box = 8 inches

∵ The width of the rectangular box = 3 inches

∵ The height of the rectangular box = 5 inches

∵ Length and width are perpendicular to each other

∴ The Δ whose legs are 3 and 8 is a right triangle

In the right Δ whose legs are 3 and 8

∵ (hypotenuse)² = (3)² + (8)²

∴ (hypotenuse)² = 9 + 64

∴ (hypotenuse)² = 73

- Take √ for both sides

∴ hypotenuse = $\sqrt{73}$ = 8.544003745

- Round it to the nearest tenth of one inch

∴ hypotenuse = 8.5 inches

The 3-dimensional diagonal is the hypotenuse of a right triangle whose legs are the vertical edge and the hypotenuse of the right triangle whose legs are 3 and 8

∵ The hypotenuse of the right triangle whose legs are 3 and

8 is 8.5 inches

∴ The length of the second leg is 8.5 inches

In the right triangle whose hypotenuse is the 3-dimensional diagonal and legs are the vertical edge , the hypotenuse of the right triangle whose legs are 3 and 8

∵ (3-dimensional diagonal)² = (5)² + (73)²

∴ (3-dimensional diagonal)² = 25 + 73

∴ (3-dimensional diagonal)² = 98

- Take √ for both sides

∴ 3-dimensional diagonal = $\sqrt{98}$ = 9.899494937

- Round it to the nearest tenth of an inch

∴ 3-dimensional diagonal = 9.9 inches

∴ The length of the three-dimensional diagonal is 9.9 inches

<em>V.I.N: you can find the length of the three-dimensional diagonal by using this rule → </em> $d=\sqrt{l^{2}+w^{2}+h^{2}}$ <em> </em>

3 0

4 years ago