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3 years ago

Last week you worked 20 hours, and earned $255.

Mathematics

1 answer:

Marrrta [24]3 years ago

6 0

Answer:

Answer:$12.75

Answer:$12.75Step-by-step explanation:

The hourly pay rate is $12.75 because all you have to do is divide $255 from 20 and you get $12.75

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Lines k and n intersect on the y-axis

avanturin [10]

a) The equation of line k is:

$y = -\frac{202}{167}x + \frac{598}{167}$

b) The equation of line j is:

$y = \frac{167}{202}x + \frac{1546}{202}$

The equation of a line, in <u>slope-intercept formula</u>, is given by:

$y = mx + b$

In which:

m is the slope, which is the rate of change.
b is the y-intercept, which is the value of y when x = 0.

Item a:

Line k intersects line m with an angle of 109º, thus:

$\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}$

In which $m_1$ and $m_2$ are the slopes of <u>k and m.</u>

Line k goes through points (-3,-1) and (5,2), thus, it's slope is:

$m_1 = \frac{2 - (-1)}{5 - (-3)} = \frac{3}{8}$

The tangent of 109 degrees is $\tan{109^{\circ}} = -\frac{29}{10}$
Thus, the slope of line m is found solving the following equation:

$\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}$

$-\frac{29}{10} = \frac{m_2 - \frac{3}{8}}{1 + \frac{3}{8}m_2}$

$m_2 - \frac{3}{8} = -\frac{29}{10} - \frac{87}{80}m_2$

$m_2 + \frac{87}{80}m_2 = -\frac{29}{10} + \frac{3}{8}$

$\frac{167m_2}{80} = \frac{-202}{80}$

$m_2 = -\frac{202}{167}$

Thus:

$y = -\frac{202}{167}x + b$

It goes through point (-2,6), that is, when $x = -2, y = 6$ , and this is used to find b.

$y = -\frac{202}{167}x + b$

$6 = -\frac{202}{167}(-2) + b$

$b = 6 - \frac{404}{167}$

$b = \frac{6(167)-404}{167}$

$b = \frac{598}{167}$

Thus. the equation of line k, in slope-intercept formula, is:

$y = -\frac{202}{167}x + \frac{598}{167}$

Item b:

Lines j and k intersect at an angle of 90º, thus they are perpendicular, which means that the multiplication of their slopes is -1.

Thus, the slope of line j is:

$-\frac{202}{167}m = -1$

$m = \frac{167}{202}$

Then

$y = \frac{167}{202}x + b$

Also goes through point (-2,6), thus:

$6 = \frac{167}{202}(-2) + b$

$b = \frac{(2)167 + 202(6)}{202}$

$b = \frac{1546}{202}$

The equation of line j is:

$y = \frac{167}{202}x + \frac{1546}{202}$

A similar problem is given at brainly.com/question/16302622

7 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

Shari had six dimes and nine nickels.Did she have enough money to buy a $0.99 pack of gum

Flura [38]

Answer:yes she does cauce she has 1.05

Step-by-step explanation:

8 0

3 years ago

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If f(x)=7x what is the inverse of f(x)

Brums [2.3K]

Answer:

f(x)=x/7

Step-by-step explanation:

f(x)=7x

y=7x

x=7y (switch y and x for inverse)

x=7y (isolate y)

x/7=y

The inverse of f(x)=7x is f(x)=x/7

8 0

3 years ago

If angles G and H are supplementary angles, and angle G is 60°, What measurement, in degrees, is angle H?

enyata [817]

Answer:

120 degrees = H

Step-by-step explanation:

Two angles whose sum is equal to 180 degrees are called supplementary angles. do 180-60 to get 120

hope it helps! :)

8 0

3 years ago

Read 2 more answers