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 Fill in the blanks to make a quadratic whose roots are 5 and -5.   x^2 + ___ x + ___  

Since the roots are +5 and –5, the quadratic factors to (x – 5)(x + 5)   

Multplied out, (x – 5)(x + 5) is x² – 25.  

Fill in the blanks with 0 and –25 ............. x²  +  0x  +  –25    

^.

https://web2.0calc.com/questions/a-circle-lies-inside-a-quarter-circle-as-shown-below

Ceiling [ 15 * 60 *1 /  (.3048^2) / 400] =  25 gallons

f(1)  = -1

f(2)  = 0

...

f(1000) = 0

Sum  = -1

\(\frac{x^2}{x - 1 + x^3} for x > 1 \)

The graph here shows that there is  no min for any x

https://www.desmos.com/calculator/jjazrygiar

The graph will approach 0 as x →  infinity

https://web2.0calc.com/questions/help-with-geometry_22606

W               A    15      X

                                   x

                              60 C

                                  3x

Z             B      20       Y

Let ACX = Angle A

Let BCY = Angle B

A + B + 60  =  180

A + B =  120 

tan ACX =   15/x

tan BCY  =  20 / (3x)

tan (A + B)  =  tan (120)

[ tan A  + tan B] /  1 - tanAtanB  =  -sqrt 3

[ 15/x + 20/(3x) ] / [ 1 - (15/x)(20/(3x)) ]  = -sqrt 3

[ 15 (3x) + 20x ]  / [ 3x^2 ]  

_____________________    =  -sqrt (3)

1 - 300 / (3x^2 )

[ 45x + 20x ] / [3x^2]

________________  =   -sqrt 3

[3x^2 - 300] / [3x^2]

65x =  sqrt (3) [300 - 3x^2]

3sqrt (3)x^2 + 65x - 300sqrt (3)  =  0

Using the Quad Formula and taking the positive value of x ≈ 5.5403

3x = 16.6209

AC  = sqrt [  15^2  + (5.5403)^2 ]  ≈ sqrt [ 225.69 ] 

BC = sqrt [ 20^2  + (16.6209)^2 ]  ≈ sqrt [676.25]

Law of Cosines

AB^2  =  AC^2 + BC^2 - 2 (AC * BC) cos (60)

AB^2  = 225.69 + 676.25 - 2 sqrt (225.69*676.25 ] * (1/2)

AB^2 = 225.69 + 676.25  - sqrt [ 225.69* 676.25 ]

AB^2  = 511.270

AB  = sqrt (511.270)  ≈  22.6

https://web2.0calc.com/questions/help-with-algebra_96

WX * YZ = 2 * 3  = 6 

We can solve the problem as follows.

Similar Triangles:

Since AB = AC and altitude AH intersects BC at D, we can split triangle ABC into two congruent right triangles (ABH and ACH).

Pythagorean Theorem in ABH:

We are given AH = 11 and DH = 5. Since these sides form a right angle (angle AHB = 90°), we can use the Pythagorean Theorem:

AB^2 = AH^2 + HD^2

AB^2 = 11^2 + 5^2

AB^2 = 121 + 25

AB^2 = 146

Taking the square root of both sides (remembering AB could be positive or negative, but the area is based on the absolute value):

AB = √146 (positive square root for finding side length)

Area of Triangle ABC:

Since triangle ABC is formed by two congruent right triangles (ABH and ACH), their areas will be equal.

Therefore, the area of triangle ABC is simply twice the area of triangle ABH.

Area of triangle ABH = 1/2 * base * height = 1/2 * AB * AH

Substitute Known Values and Simplify:

We know AH = 11 and from step 2, AB = √146. Substitute these values:

Area of triangle ABC = 2 * (1/2 * √146 * 11)

Therefore, the area of triangle ABC is 11*sqrt(146).

https://web2.0calc.com/questions/pascal-s-triangle_25

https://web2.0calc.com/questions/rates_51

In a row of five squares, each square is to be colored either red, yellow, or blue, so that no two consecutive squares have the same color. How many ways are there to color the five squares, if there must be at least three yellow squares?  

In order not to have two yellows next to each other,  

the three yellows must occupy squares 1, 3, and 5. 

That leaves squares 2 and 4 to fill, and two colors  

to do it with. There are only four ways. 

Blue Blue     Red Red     Blue Red     Red Blue  

So, the answer is four ways.  

_.

To find the number of positive integers that satisfy both conditions, we can approach it systematically by considering the possible combinations of digits that sum up to 5. Since each digit can be 1, 2, or 3, and we need to find the combinations that sum up to 5, we can start with the smallest possible digit combinations and incrementally increase them until we reach the sum of 5.

Here are the possible combinations:

(1, 1, 1, 1, 1)

(1, 1, 1, 2)

(1, 1, 2, 1)

(1, 2, 1, 1)

(2, 1, 1, 1)

(1, 1, 3)

(1, 3, 1)

(3, 1, 1)

(1, 2, 2)

(2, 1, 2)

(2, 2, 1)

(2, 3)

(3, 2).

\(A(a,a^2)\\ B(b,b^2)\\ \dfrac{b^2-a^2}{b-a}=2\\ \dfrac{(b+a)(b-a)}{b-a}=2\)

\(\color{blue}b+a=2\\ \)

 !

2.   H is the  orthocenter of the triangle  Draw CH  and extend it  to F where CF is the third altitude of the  triangle

Angle AHE  =  180   - 128 =  52

Angle AEH  = 90

Angle HAE  =  90  - 52 = 38

Angle BAH  =  28

Angle BAC =  Angle BAH + Angle HAE  =  28 + 38  = 66 

Angle CFA =  90

Angle BAC =  66

Angle FCA =  180 - 90 - 66   = 24

But Angle FCA = Angle HCA =  24

The positive integers are, 

11111, 1112, 1121, 1211, 2111, 221, 212, 122, 311, 131, 113, 32, 23.

So, there are 13 numbers that satisfy the condition.

Set the center of the small circle to be O₁, and the center of the larger circle to be O₂.

Draw a perpendicular line passing through O₁, perpendicular to BO₂, label the foot of the perpendicular point C.

ABCO₁ is a rectangle, and so AB = CO₁.

Also, CO₂ = 4 - 1 = 3

Apply the pythagorean theorem, so (CO₁)²=(O₁O₂)²-(CO₂)²=25 - 9=16.

Therefore, CO₁=AB=4.

https://web2.0calc.com/questions/help-coordinates_136

11111          

1112   1121  1211  2111

122      212    221

113      131    311

 23         32

https://web2.0calc.com/questions/help-with-coordinates_36

Here's a very similar problem answered by hectictar

https://web2.0calc.com/questions/help-asap-tysm_1

https://web2.0calc.com/questions/circles_147