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3 years ago

Find the x of (2x+8)+(71-x)=180

Mathematics

2 answers:

fomenos3 years ago

8 0

Answer:

x=100

Step-by-step explanation:

(2x+8)+(71-x)=180

2x+8+71-x=180

x+80=180

x=180-80

x=100

ollegr [7]3 years ago

5 0

Answer:

x = 101

Step-by-step explanation:

First, simplify the equation

2x + 8 + 71 - x = 180

x + 79 = 180

both side we subtract with 79 so the 79 on the left go away

x = 180 - 79

x = 101

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This so unfaired. dis dude be like:

amm1812

Answer:

correct

Step-by-step explanation:

pls mak as brainlyest if it is unfair

it is so unfair

4 0

2 years ago

Read 2 more answers

Factor 28+56t+28w28+56t+28w28, plus, 56, t, plus, 28, w to identify the equivalent expressions. Choose 2 answers:

Effectus [21]

Answer: The factorization of $28+56t+28w$ $=28(1+2t+w)$

The factors of $28+56t+28w$ are 28 and 1+2t+w.

Step-by-step explanation:

The given expression : $28+56t+28w$

To factorize it, we need to find the common factor.

As 56 can be written as 2 x 28.

So, the above expression would become

$28+2\times28t+28w$

Now, taking 28 as common from all the terms, we will get

$28(1+2t+w)$

Thus, the factorization of $28+56t+28w$ $=28(1+2t+w)$

And the factors of $28+56t+28w$ are 28 and 1+2t+w.

6 0

2 years ago

Andrew must spend less than 53$ on meals during the weekend. he has already spent 21$ on meals costing 8$ average. how many addi

Shkiper50 [21]

Andrew can buy maximum 3 meals this weekend.

From given question,

Andrew must spend less than 53$ on meals during the weekend.

He has already spent 21$ on meals costing 8$ average.

Let x, the number of meals

So, we get an inequality,

8x + 21 < 53

We need to find the number of meals he can buy this weekend.

From above inequality,

⇒ 8x + 21 < 53

⇒ 8x < 53 - 21

⇒ 8x < 32

⇒ x < 4

This means, from 1 to 3 meals.

Therefore, Andrew can buy maximum 3 meals this weekend.

Learn more about an inequality here:

brainly.com/question/19003099

#SPJ4

3 0

1 year ago

5x+2y=22. -2x+6y=3 What is the X coordinate of the solution. Round to the nearest tenth

just olya [345]

Multiply first example by 3 and subtract by second example, you will get the following:

3*(5x + 2y) = 3*22 => 3*5x + 3*2y = 66 => 15x + 6y = 66
now subtraction:
(15x + 6y) - (-2x + 6y) = 66 - 3
15x + 6y + 2x - 6y = 63
15x + 2x + 6y - 6y = 63
17x = 63
x = 63/17 ≈ 3.705..... and as it says to round to the nearest tenth our answer would be: x = 3.7

8 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