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larisa [96]
2 years ago
7

First to answer correctly will be marked brainliest 40 points as well

Mathematics
1 answer:
skelet666 [1.2K]2 years ago
7 0

Answer:

see explanation↓

Step-by-step explanation:

(a)

(3x+2)-(x+5)=3x+2-x-5

(3x+2)-(x+5)=2x-3

ANSWER: Linear binomial

(b)

(3x+8)+(4x^{2}-x)=3x+8+4x^{2}-x

(3x+8)+(4x^{2}-x)=4x^{2} +2x+8

ANSWER: Quadratic trinomial

(c)

(x^{2}+7x)-(x^{2}+5)=x^{2}+7x-x^{2}-5

(x^2+7x)-(x^2+5)=7x-5

ANSWER: Linear binomial

(d)

(2x+8)+(3x^2-2)=2x+8+3x^2-2

(2x+8)+(3x^2-2)=3x^2+2x+6

ANSWER: Quadratic trinomial

(e)

(x^2+5x-2)+(8-5x)=x^2+5x-2+8-5x

(x^2+5x-2)+(8-5x)=x^2+6

ANSWER: Quadratic binomial



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M(d)=-0.123x^2+148.83x-21.07<br> When will it reach maximum height of 45000 feet
Setler [38]

Answer: time to reach maximum height = 605

Step-by-step explanation:

The expression that relates height with time is

M(d)=-0.123x^2+148.83x-21.07

Where time = x

d = height

The expression is a quadratic equation. If height of the object is plotted against time, the resulting graph is a parabola whose vertex is equal to the maximum height attained by the object. The time, x corresponding to the vertex is the time for it to reach maximum height.

To find x,

x = -b/2a

Where a= 0.123

b = 148.83

x = - 148.83 / -2 × -0.123

x = -148.83 / -0.246

= 605

Substituting x = 605 in the equation,

M(d)=-0.123x^2+148.83x-21.07

d = -0.123 × 605^2 + 148.83×605

= - 45021.075+ 90042.15

= 45020.4

Approximately 45000 feet

6 0
3 years ago
Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity
kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

< at^2+1,t>\cdot < 2at,1>=

                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




6 0
3 years ago
100 POINTS + BRAINLIEST !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!11<br> Solve x³ - 9x^(1.5) + 8 = 0
VLD [36.1K]

Answer:

x=1, x=4

Step-by-step explanation:

3 0
2 years ago
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Dmitriy789 [7]
The letters in MATH can be arranged 16 ways. D.
6 0
3 years ago
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Pleaee help me with this im stuck
AURORKA [14]

the answers are clear if you use priority and equation rules.

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3 years ago
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