Y = 3 - 5x solve for x

Ill give u branliest if u help plsss The graphs of f (x) and g() are shown.

Solnce55 [7]

Answer:

the Answer is 2 hope that helps

5 0

2 years ago

HELP PLEASE I NEED THIS FOR MY TEST TODAY HELP

maw [93]

Answer:

j is the answer hope i got it in time

4 0

2 years ago

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If a graph shows the function y=3x what are the coordinates of the intercepts

vladimir2022 [97]

Answer:

graph{3x+5 [-10, 10, -5, 5]}

x

intercept:

x

=

−

5

3

y

intercept:

y

=

5

Explanation:

For a linear graph, the quickest way to sketch the function is to determine the

x

and

y

intercepts and draw a line between the two: this line is our graph.

Let's calculate the

y

intercept first:

With any function,

y

intercepts where

x

=

0

.

Therefore, substituting

x

=

0

into the equation, we get:

y

=

3

⋅

0

+

5

y

=

5

Therefore, the

y

intercept cuts through the point (0,5)

Let's calculate the

x

intercept next:

Recall that with any function:

y

intercepts where

x

=

0

.

The opposite is also true: with any function

x

intercepts where

y

=

0

.

If we substitute

y

=

0

, we get:

0

=

3

x

+

5

Let's now rearrange and solve for

x

to calculate the

x

intercept.

−

5

=

3

x

−

5

3

=

x

Therefore, the

x

intercept cuts through the point

(

−

5

3

,

0

)

.

Now we have both the

x

and

y

intercepts, all we have to do is essentially plot both intercepts on a set of axis and draw a line between them

The graph of the function

y

=

3

x

+

5

:

graph{3x+5 [-10, 10, -5, 5]}

3 0

3 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

On Monday, a bakery sold 67 dozen cookies. How many total cookies did the bakery sell?

gulaghasi [49]

Answer:

804

Step-by-step explanation:

dozen means 12

67 times 12 is 804

8 0

1 year ago

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