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3 years ago

Rezolvati 1) -2x+1= -7 2) 6+3x= -54+x 3) 2x+9= 7 +x 4) 3-2x= -x -6 5) -3(x+1) = -6 6) -4 (x -2)=12 7) -7(x-3)= -14

Mathematics

1 answer:

Natalija [7]3 years ago

3 0

<span>Răspunsurile sunt

1. x= 4
2. x = -30
3. x = -2
4. x = 9
5. x = 1
6. x = -1
7. x = 5
</span>

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Consider the function f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0

dybincka [34]

I suppose you mean

$f(x)=\cos(x^2)$

Recall that

$\cos x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}$

which converges everywhere. Then by substitution,

$\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}$

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

$f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}$

(starting at $n=1$ because the summand is 0 when $n=0$ )

b. Naturally, the differentiated series represents

$f'(x)=-2x\sin(x^2)$

To see this, recalling the series for $\sin x$ , we know

$\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}$

Multiplying by $-2x$ gives

$-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}$

and from here,

$-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}$

$-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)$

c. This series also converges everywhere. By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}$

The limit is 0, so any choice of $x$ satisfies the convergence condition.

3 0

3 years ago

Which expression is equivalent to 15p^-4q^-6/-20p^-12q^-3 in simplified from? assume p=/ 0,q=/0

enyata [817]

(15p⁺⁴.q⁻⁶) /(-20p⁻¹².q⁻³).

Remember that a⁻ⁿ = 1/aⁿ and 1/a⁻ⁿ = aⁿ

(-15/4).(p⁻⁴.q⁻⁶)(p⁺¹².q⁺³).

(-15/20).(p⁻⁴.p¹².q⁻⁶.q³)

Remember aⁿ.aˣ = aⁿ⁺ˣ

(-15/20).(p⁸.q⁻³)

-3/5(p⁸.q⁻³)

3 0

3 years ago

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In the diagram, TC represents a vertical building. The points, A and B, are on the same level as the foot C of the building such

Alenkinab [10]

Answer:

(a) The height of the building is 60.06 m

(b) The distance AB is 139.43 m

Step-by-step explanation:

The given parameters are

Given that segment BT = segment AT + 29

By trigonometric ratios, we have;

cos∠ATC = CT/AT

cos∠BTC = CT/BT

Therefore, we have;

cos(40°) = CT/AT.................................(1)

cos(56°) = CT/BT = CT/(AT + 29).....(2)

cos(56°) = CT/(AT + 29)......................(3)

From equation (1)

CT = AT×cos(40°)

From equation (3)

AT×cos(56°) + 29 × cos(56°) = CT

Therefore;

AT×cos(40°) = AT×cos(56°) + 29 × cos(56°)

AT×cos(40°) - AT×cos(56°) = 29 × cos(56°)

AT×(cos(40°) - cos(56°)) = 29 × cos(56°)

AT = 29 × cos(56°)/(cos(40°) - cos(56°)) = 78.4 m

TC = CT = AT×cos(40°) = 78.4×cos(40°) = 60.06 m

The height of the building = 60.06 m

(b) BT = AT + 29 = 78.4 m + 29 m= 107.4 m

AB = AT×sin(∠ATC ) + BT×sin(∠BTC) = 78.4×sin(40°) + 107.4×sin(56°) = 139.43 m

The distance AB = 139.43 m.

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2 years ago

(5)^2 - (-9)^2 / (5) - (-9)

timama [110]

89/5
Alternative form

1) 17.8
2)17 4/5

4 0

2 years ago

A triangle has two sides with lengths 7 m and 12 m. Which of the following lengths could represent the length of the third side?

hichkok12 [17]

B is the correct answer because in order to form a triangle the sum of each two sides must be greater than the third side

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3 years ago

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