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3 years ago

The sum of 13 and twice a number

1 answer:

7 0

This would be 13+2x because if x is the number, then 2 times it would be 2x and the sum means addition so it would be 13+2x

You might be interested in

Find the square root of the following using division method . iv) 36360

sveticcg [70]

Answer:

190.684

Step-by-step explanation:

To find the square root of 36360, we place a bar over the numbers that we want to find their square root in pairs

, 1 $\dot{3}$ $\overline{63}$ $\overline{60}$

, 1 1

2, 9 2 $\overline{63}$

9 261

38, 0 2 60

0 0

380, 6 26000

, 6 22800

3806, 8 320000

, 8 304480

38068, 4 1552000

Which gives 190.684

3 63 60

We select the divisor to be the largest number which when squares will be equal to the largest number on the left which is 1

We have 1 * 1 = 1

We subtract it from the 3 to get 2

We add the 1's to get 2

We bring down the next pair which is 63 to form the next dividend, 263

We look for a number which will multiply itself to get 263, which is 9, we then have 29 × 9 = 261

We continue till we arrive at the vertical paired numbers which is 190.684.

7 0

3 years ago

Suppose that $5500 is placed in an account that pays 2% interest compounded each year.

pogonyaev

Answer: After 1 year: $5,610

After 2 years: $5,722.20

Step-by-step explanation: Use the formula for periodic compounding interest, which is

A = P(1 + r/n)^(nt), where A is the final amount, P is the initial deposit, r is the interest rate as a decimal, n is the number of times the interest is compounded per year, and t is how many years.

Here, P = 5,500, r = 0.02 (that's 2% as a decimal), n = 1,

t = 1 for the first answer, t = 2 for the second answer (1 year, then for 2 years)

Plug the known values in to solve...

For 1 year...

A = 5,500(1 + 0.02/1)^(1*1)

A = 5,500(1.02)^1

A = 5,610

For 2 years...

A = 5,500(1 + 0.02/1)^(1*2)

A = 5,500(1.02)²

A = 5,722.20

3 0

2 years ago

The sun appears to move across the sky, because the earth spins on its axis. To a person standing on the earth, the sun subtends

Liono4ka [1.6K]

Answer:

128 seconds

Step-by-step explanation:

The sun subtends an angle $\theta=9.28\times10^{-3}$ rad

Angular velocity equation is $\omega = \frac{\theta}{t}$

where $\omega$ is angular velocity and t is time.

Earth spins at a rate of :

$\omega = \frac{2 \pi rad}{24 h \times \frac{3600 s}{1 h}} = 7.27221\times10^{-5}$ rad/s

Isolating time (t) from angular velocity equation gives:

$\omega = \frac{\theta}{t}$

$t = \frac{\theta}{\omega}$

$t = \frac{9.28\times10^{-3}}{7.27221\times10^{-5}}$

$t = 128 s$

8 0

3 years ago

What is the reciprocal of -5.2/2.5 ? need an answer fast please

V125BC [204]

Answer:

− 0.48076923

Step-by-step explanation:

4 0

2 years ago

The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for

4vir4ik [10]

A) $g=f'$ is continuous, so $f$ is also continuous. This means if we were to integrate $g$ , the same constant of integration would apply across its entire domain. Over $0$ , we have $g(x)=2x$ . This means that

$f_{0$

For $f$ to be continuous, we need the limit as $x\to1^-$ to match $f(1)=3$ . This means we must have

$\displaystyle\lim_{x\to1}x^2+C=1+C=3\implies C=2$

Now, over $x$ , we have $g(x)=-3$ , so $f_{x$ , which means $f(-5)=17$ .

b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,

$\displaystyle\int_1^6g(x)=4+\int_3^62(x-4)^2\,\mathrm dx=4+6=10$

c) $f$ is increasing when $f'=g>0$ , and concave upward when $f''=g'>0$ , i.e. when $g$ is also increasing.

We have $g>0$ over the intervals $0$ and $x>4$ . We can additionally see that $g'>0$ only on $0$ and $x>4$ .

d) Inflection points occur when $f''=g'=0$ , and at such a point, to either side the sign of the second derivative $f''=g'$ changes. We see this happening at $x=4$ , for which $g'=0$ , and to the left of $x=4$ we have $g$ decreasing, then increasing along the other side.

We also have $g'=0$ along the interval $-1$ , but even if we were to allow an entire interval as a "site of inflection", we can see that $g'>0$ to either side, so concavity would not change.

5 0

2 years ago