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2 years ago

Evaluate the expression -3+2× if the value of x = -2 i got answer -7

Mathematics

2 answers:

seraphim [82]2 years ago

4 0

Answer:

-7

Yep you were right!

Step-by-step explanation:

-3+2(-2)

-3+-4

-7

nikitadnepr [17]2 years ago

4 0

Answer:

-7

Step-by-step explanation:

The value of -3 + 2x at x = -2 is -3 + 2(-2) = -3 - 4 = -7. You are correct.

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Question 17

SOVA2 [1]

Answer:

1.8 × 10^10

Step-by-step explanation:

(7.2 × 10 = 72)

(72) ÷ (4 × 10^-9) = 1.8 × 10^10

3 0

2 years ago

Convert Y = -7/2x-11 to Standard Form

monitta

Answer:

7x + 2y = -22

Step-by-step explanation:

Standard form looks like Ax + By = C.

Start with Y = -7/2x-11. Better to write that as Y = (-7/2)x - 11 to emphasize that the coefficient of x is the fraction -7/2.

Move the (-7/2)x term to the left:

(7/2)x + y = -11

Multiply all terms by 2 to eliminate the fractions:

7x + 2y = -22 (answer)

6 0

3 years ago

The top of a ladder rests at a height of 15 feet against the side of a house. If the base of the ladder is 6 feet from the house

Ksju [112]

90
15 x 6 = 90
You need to multiply

6 0

3 years ago

Read 2 more answers

8, 2, 0, 2, 8, 18<br> What are the next three numbers in the sequence?<br> What is the pattern?

Tom [10]

Answer:

22,40,62

Step-by-step explanation:

0+2=2 2+6=8 8+10=18

2 6 10 14 18 22 four in between

so you will add these numbers each time

18+14=22 22+18=40 40+22=62

4 0

2 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago