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2 years ago

Combine the like terms to create an equivalent expression. 5 + 3t +4-t

Mathematics

2 answers:

marshall27 [118]2 years ago

7 0

Answer:

2t + 9

Step-by-step explanation:

5 + 4 = 9

3t - 1t = 2t

vova2212 [387]2 years ago

3 0

Answer:

2t + 9

Step-by-step explanation:

5 and 4 are like terms. 3t and -t are like terms

= 2t + 9

or 9 + 2t

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Find f such that the given conditions are satisfiedf’(x)=x-4, f(2)=-1

kicyunya [14]

Given:

$f^{\prime}\left(x\right)=x-4,\text{ and}f\left(2\right)=-1$

To find:

The correct function.

Explanation:

Let us consider the function given in option D.

$f(x)=\frac{x^2}{2}-4x+5$

Differentiating with respect to x we get,

$\begin{gathered} f^{\prime}(x)=\frac{2x}{2}-4 \\ f^{\prime}(x)=x-4 \end{gathered}$

Substituting x = 2 in the function f(x), we get

$\begin{gathered} f(2)=\frac{2^2}{2}-4(2)+5 \\ =2-8+5 \\ =-6+5 \\ f(2)=-1 \end{gathered}$

Therefore, the given conditions are satisfied.

So, the function is,

$f(x)=\frac{x^{2}}{2}-4x+5$

Final answer: Option D

6 0

1 year ago

You want to put shingles on the outside walls and solar panel the roof of the barn shown. It costs $2.50 for each square meter

Luden [163]

Answer: 975

Step-by-step explanation:

The cost would be 975

3 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

3 years ago

Find the value of x in this figure

Serjik [45]

Answer:

here question was not clear

4 0

3 years ago

Read 2 more answers

Solve for x<br> Write the smaller solution first, and the larger solution second.<br> -x^2-8=-33

kenny6666 [7]

Answer:

x = -5, 5

Step-by-step explanation:

-x^2-8=-33

Add 8 to each side

-x^2-8+8=-33+8

-x^2 = -25

Divide each side by -1

x^2 = 25

Take the square root of each side

sqrt(x^2) = ±sqrt(25)

x = ±5

8 0

3 years ago

Read 2 more answers