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3 years ago

76 + 4= add the ones and the tens. Do you regroup? yes or no

Mathematics

1 answer:

nikklg [1K]3 years ago

6 0

76 + 4 is 80... don't know what you're asking

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Newton’s Second Law of Motion states that the acceleration of an object is directly proportional to the force exerted on the obj

aev [14]

The force the same object will exert if it accelerates at a rate of 7 m/s² is 35N

From the question,

We are to determine the force the object will exert.

From the information,

Newton’s Second Law of Motion states that the acceleration of an object is directly proportional to the force exerted on the object

Let acceleration be a

and

Force be F

Then we can write that,

F ∝ a

Then,

Introducing the constant, mass, m

We get

F = ma

Now, we will determine the mass of the object

F = 15 N

a = 3 m/s²

Putting the parameters into the formula, we get

15 = m × 3

∴ m = 15 ÷ 3

m = 5 kg

The mass of the object is 5 kg.

Now, to determine the force the same object will exert if it accelerates at a rate of 7 m/s²

That is,

a = 7 m/s²

and

m = 5 kg

Putting the parameters into the formula, we get

F = 5 × 7

F = 35 N

Hence, the force the same object will exert if it accelerates at a rate of 7 m/s² is 35N

Learn more here: brainly.com/question/13590154

4 0

2 years ago

Given the triangle below, what is the length of the third side, rounded to the nearest whole number?

Leno4ka [110]

Answer:

Step-by-step explanation:

You need the Law of Cosines for this, namely:

$x^2=21^2+14^2-2(21)(14)cos58$ where x is the missing side.

$x^2=441+196-311.5925$ and

$x^2=325.4075$ so

x = 18.0 or just 18

7 0

3 years ago

Solve for x. -x/3=2 x= -6 x= -1 x = 6

LenKa [72]

Answer:

x = -6

Step-by-step explanation:

Solve for x:

-x/3 = 2

Multiply both sides of -x/3 = 2 by -3:

(-3 (-x))/3 = -3×2

-3×(-1)/3 = (-3 (-1))/3:

(-3 (-1))/3 x = -3×2

(-3)/3 = (3 (-1))/3 = -1:

--1 x = -3×2

(-1)^2 = 1:

x = -3×2

-3×2 = -6:

Answer: x = -6

6 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

3 years ago

4. 110 in. = _ ft_ in.

aliina [53]

Answer:

110 in. = 9 1/6 ft, or 9 ft 2 in

Step-by-step explanation:

Please try to do a better job of formatting your next question. Thanks.

110 in 1 ft

-------- * --------- = 9 1/6 ft or 9 ft 2 in

1 12 in

4 0

3 years ago