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3 years ago

.. 600 is what percent of 150?

Mathematics

1 answer:

Roman55 [17]3 years ago

4 0

Answer:

400%

Step-by-step explanation:

We can write a percent proportion:

600/x = 150/100

a:b = c:d

a and d are linked and b and c are linked. So if a is missing, you would do (b*c)/d to get a, if b was missing, you would do (a*d)/c to get b, etc.

So over here we do

(600*100)/150 = 60,000/150 = 400, so x = 400, so that means it's 400% of 150

We can also use benchmark fractions which I find much easier:

100% of 150 is 150.

So 200% of 150 is 300

So 300% of 150 is 450

And 400% of 140 is 600

Every time it increases by 100%, it also increases by 150 on the other side, which makes sense

Let me know if you have any questions.

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3 years ago

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almond37 [142]

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Find the circumference of a circle with a radius of 18.5m, give answers in pie

Simora [160]

Answer:

C = pi * 37m

Step-by-step explanation:

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3 years ago

Question attached plz answer

Bezzdna [24]

Answer:

x = - 2.5

Step-by-step explanation:

Given that the sketch represents

y = x² + bx + c

The graph crosses the y- axis at (0 , - 14), thus c = - 14

y = x² + bx - 14

Given the graph crosses the x- axis at (2, 0), then

0 = 2² + 2b - 14

0 = 4 + 2b - 14 = 2b - 10 ( add 10 to both sides )

10 = 2b ( divide both sides by 2 )

b = 5

y = x² + 5x - 14 ← represents the graph

let y = 0 , then

x² + 5x - 14 = 0 ← in standard form

(x + 7)(x - 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 7 = 0 ⇒ x = - 7

x - 2 = 0 ⇒ x = 2

The x- intercepts are x = - 7 and x = 2

The vertex lies on the axis of symmetry which is midway between the x- intercepts, thus

the x- coordinate of the turning point is $\frac{-7+2}{2}$ = $\frac{-5}{2}$ = - 2.5

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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

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

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

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3 years ago