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Mashutka [201]
3 years ago
7

="absmiddle" class="latex-formula">
Mathematics
1 answer:
Debora [2.8K]3 years ago
4 0
The answer is x= 1/4
5x2+2x-3=0
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Does anyone know all the concepts/rules needed to learn math in college? I want to make sure I have things I need now in my seni
Mandarinka [93]

Answer:

Basic college math topics include whole numbers, fractions, numbers, decimals and integers. Problem solving, algebra, percents and geometry are also included in course content.

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
How many terms are in the expression <br> 5x^3-8x^2y+4xy-4
bazaltina [42]

Answer:

<h2>4 terms</h2>

Step-by-step explanation:

Term 1: 5x^3

Term 2: 8x^2y

Term 3: 4xy

Term 4: 4

8 0
3 years ago
Given: KLMN is a trapezoid m∠N = m∠KML ME ⊥ KN , ME = 3 5 , KE = 8, LM KN = 3 5 Find: KM, LM, KN, Area of KLMN
prisoha [69]
A) Find KM∠KEM is a right angle hence ΔKEM is a right angled triangle Using Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides we can answer the 
KM² = KE² + ME²KM² = 8² + (3√5)²       = 64 + 9x5KM = √109KM = 10.44
b)Find LMThe ratio of LM:KN is 3:5 hence if we take the length of one unit as xlength of LM is 3xand the length of KN is 5x ∠K and ∠N are equal making it a isosceles trapezoid. A line from L that cuts KN perpendicularly at D makes KE = DN
KN = LM + 2x 2x = KE + DN2x = 8+8x = 8LM = 3x = 3*8 = 24
c)Find KN Since ∠K and ∠N are equal, when we take the 2 triangles KEM and LDN, they both have the same height ME = LD.
∠K = ∠N  Hence KE = DN the distance ED = LMhence KN = KE + ED + DN since ED = LM = 24and KE + DN = 16KN = 16 + 24 = 40
d)Find area KLMNArea of trapezium can be calculated using the  formula below Area = 1/2 x perpendicular height between parallel lines x (sum of the parallel sides)substituting values into the general equationArea = 1/2 * ME * (KN+ LM)          = 1/2 * 3√5 * (40 + 24)         = 1/2 * 3√5 * 64         = 3 x 2.23 * 32         = 214.66 units²
7 0
3 years ago
Slove the equation V=(20-2x)(15-2x)(x) for V
Schach [20]

Answer:

V = 300x + -70x2 + 4x3

Step-by-step explanation:

Simplifying

V = (20 + -2x)(15 + -2x)(x)

Reorder the terms for easier multiplication:

V = x(20 + -2x)(15 + -2x)

Multiply (20 + -2x) * (15 + -2x)

V = x(20(15 + -2x) + -2x * (15 + -2x))

V = x((15 * 20 + -2x * 20) + -2x * (15 + -2x))

V = x((300 + -40x) + -2x * (15 + -2x))

V = x(300 + -40x + (15 * -2x + -2x * -2x))

V = x(300 + -40x + (-30x + 4x2))

Combine like terms: -40x + -30x = -70x

V = x(300 + -70x + 4x2)

V = (300 * x + -70x * x + 4x2 * x)

V = (300x + -70x2 + 4x3)

Solving

V = 300x + -70x2 + 4x3

Solving for variable 'V'.

Move all terms containing V to the left, all other terms to the right.

Simplifying

V = 300x + -70x2 + 4x3

6 0
3 years ago
Can anyone help?i been going at this for a while and i still cant figure it out, i dont really understand how to even get the an
Feliz [49]

Answer:

A) 44.26

B) 9.43

C) (2.5,3.5)

D) OH

E) 26.26

Step-by-step explanation:

Let’s figure out the lengths first.  Three are straight line, so are easy:

HE=9, SE=11, SU=9

The other two, you need to use the distance formula or Pythagorean Theorem. OH=8²+5²=c² = 64+25=c² = 89=c² = √89 = 9.43

OU=3²+5²=c² = 9+25=c² = 34=c² = √34 = 5.83

Now we can start answering the questions.

A) 9+11+9+9.43+5.83=44.26

B) 9.43

C) You already have: (2.5,3.5)

D) OH=9.43.  EH=9.  OH is longer.

E) HU is 11. OH is 9.43.  OU is 5.83.  11+9.43+5.83=26.26

5 0
3 years ago
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