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3 years ago

Slope= 5 passing through (-4,6)

Mathematics

1 answer:

olga2289 [7]3 years ago

4 0

Answer:

y = 5x +26

Step-by-step explanation:

y - y1 = m(x - x1)

y - 6 = 5(x - -4)

y - 6 = 5(x + 4)

y - 6 = 5x +20

y = 5x + 26

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100 point please help me

lutik1710 [3]

Answer:

it says 5 points

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A truck traveled 205 miles in 3/12 hours. The distance is the product of the rate and the time. To the nearest tenth, what is th

Rufina [12.5K]

Distance = rate*time, or

distance = (average speed)* time, so

average speed = distance/ time.

average speed = 205 mi / (3/12) h = 205 mi/ (1/4)h = 205*4 mi/h=

= 820 mi/h

It is looks to high for speed of the truck,

probably time should be 3.12 h,

then

average speed = 205 mi / (3.12) h ≈ 65.71 mi/h.

That looks close to real life.

8 0

3 years ago

Divide the polynomials.

Mnenie [13.5K]

Answer:

$\dfrac{3x^3+x-11}{x+1}=3x^2-3x+4-\dfrac{15}{x+1}$

Step-by-step explanation:

We want to determine the result of the quotient: $\dfrac{3x^3+x-11}{x+1}$

We follow the procedure of long division which is set out int he table below.

$\left|\begin{array}{c|c}&3x^2-3x+4\\-----&-----\\x+1&3x^3+x-11\\Subtract&-(3x^3+3x^2)\\&------\\&-3x^2+x-11\\Subtract&-3x^2-3x\\&------\\&4x-11\\Subtract&4x+4\\&------\\&-15\end{array}\right|$

Therefore:

$\dfrac{3x^3+x-11}{x+1}=3x^2-3x+4-\dfrac{15}{x+1}$

3 0

4 years ago

Even the measure ofAngle six is 62 what is the measure of angle five

Agata [3.3K]

Step-by-step explanation:

If you look closely at this image, you can see that ∠6 and ∠5 together will be equal to line c.

A straight line is equal to 180 so if we do...

$180-62$

= 118.

<u>∠5 will be equal to 118.</u>

If we go even further, because vertical angles are congruent...

∠7 = 62

∠8 = 118

8 0

3 years ago

Read 2 more answers

Determine the exact formula for the following discrete models:

marshall27 [118]

I'm partial to solving with generating functions. Let

$T(x)=\displaystyle\sum_{n\ge0}t_nx^n$

Multiply both sides of the recurrence by $x^{n+2}$ and sum over all $n\ge0$ .

$\displaystyle\sum_{n\ge0}2t_{n+2}x^{n+2}=\sum_{n\ge0}3t_{n+1}x^{n+2}+\sum_{n\ge0}2t_nx^{n+2}$

Shift the indices and factor out powers of $x$ as needed so that each series starts at the same index and power of $x$ .

$\displaystyle2\sum_{n\ge2}2t_nx^n=3x\sum_{n\ge1}t_nx^n+2x^2\sum_{n\ge0}t_nx^n$

Now we can write each series in terms of the generating function $T(x)$ . Pull out the first few terms so that each series starts at the same index $n=0$ .

$2(T(x)-t_0-t_1x)=3x(T(x)-t_0)+2x^2T(x)$

Solve for $T(x)$ :

$T(x)=\dfrac{2-3x}{2-3x-2x^2}=\dfrac{2-3x}{(2+x)(1-2x)}$

Splitting into partial fractions gives

$T(x)=\dfrac85\dfrac1{2+x}+\dfrac15\dfrac1{1-2x}$

which we can write as geometric series,

$T(x)=\displaystyle\frac8{10}\sum_{n\ge0}\left(-\frac x2\right)^n+\frac15\sum_{n\ge0}(2x)^n$

$T(x)=\displaystyle\sum_{n\ge0}\left(\frac45\left(-\frac12\right)^n+\frac{2^n}5\right)x^n$

which tells us

$\boxed{t_n=\dfrac45\left(-\dfrac12\right)^n+\dfrac{2^n}5}$

# # #

Just to illustrate another method you could consider, you can write the second recurrence in matrix form as

$49y_{n+2}=-16y_n\implies y_{n+2}=-\dfrac{16}{49}y_n\implies\begin{bmatrix}y_{n+2}\\y_{n+1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}\begin{bmatrix}y_{n+1}\\y_n\end{bmatrix}$

By substitution, you can show that

$\begin{bmatrix}y_{n+2}\\y_{n+1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}^{n+1}\begin{bmatrix}y_1\\y_0\end{bmatrix}$

$\begin{bmatrix}y_n\\y_{n-1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}^{n-1}\begin{bmatrix}y_1\\y_0\end{bmatrix}$

Then solving the recurrence is a matter of diagonalizing the coefficient matrix, raising to the power of $n-1$ , then multiplying by the column vector containing the initial values. The solution itself would be the entry in the first row of the resulting matrix.

5 0

3 years ago