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3 years ago

Emily gives her bird 1/8 cup of bird seeds along with 1/3 cup of grains each day. How much food is given each day to the bird

Mathematics

1 answer:

ANEK [815]3 years ago

6 0

Answer:

11/24 cup

Step-by-step explanation:

What you would do is add 1/8 plus 1/3. To add two fractions, each much have the same denominator. In this case, you would need to multiply 1/8 into 3/24 by multiplying 1/8 by 3 on the denominator and numerator. For 1/3, multiply each the numerator and denominator by 8 getting the fraction 8/24. Now you can line up each fraction and add the numerators, keeping the denominator at 24. That answer would be 11/24. This cannot be simplified, so the answer is 11/24 cup of food.

Whenever you are confused with assignments, it is best to tell your parent to help you or teacher the next day. Hope this helps though!

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<img src="https://tex.z-dn.net/?f=5x%20-%2010%20%5Cleqslant%2020" id="TexFormula1" title="5x - 10 \leqslant 20" alt="5x - 10 \le

OlgaM077 [116]

The answer is x = 6
Explanation:
First, you must assume that 5x-10=20. Next, you add 10 to 20, which makes the equation 5x=30. Next, divide 30 by 5, which then makes the equation x=6. When you substitute 6 in the original equation, it works, because 5(6)-10 equals 20.

6 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

2 years ago

Three consecutive odd integers are such that the square of the third integer is 1515 greater than greater than the sum of the sq

Ksju [112]

Let
x-----------> first <span>odd integer
x+2--------> second consecutive odd integer
x+4-------> third consecutive odd integer

we know that
(x+4)</span>²=15+x²+(x+2)²-------> x²+8x+16=15+x²+x²+4x+4
x²+8x+16=19+2x²+4x-------> x²-4x+3
x²-4x+3=0

using a graph tool----------> to calculate the quadratic equation
see the attached figure
the solution is
x=1
x=3

the answer is
the first odd integer x is 1
the second consecutive odd integer x+2 is 3
the third consecutive odd integer x+4 is 5

7 0

3 years ago

Write a recursive formula for an, the nth term of the sequence 3, -4, -11

yKpoI14uk [10]

Answer:

, 12, 48, 192...

a. Write a recursive formula for the nth term of the sequence

Ans: a(n+1) = 4*a(n)

------------------------------

b. Write a general formula for the nth term of the sequence

a(n) = 3*4^(n-1)

--------------

c. Calculate S10 for this sequence

Geometric sequence with a(1) = 3 and r = 4

----?

Step-by-step explanation:

8 0

2 years ago

What is the length of the missing leg in this right triangle?18 mm 24 mm 26 mm 32 mm

topjm [15]

Answer:

24 mm

Step-by-step explanation:

25^2 - 7^2

625 - 49

= 576

missing side = √576 = 24

8 0

3 years ago

Read 2 more answers