Share £880 into the ratio of 3:1

Lilit [14]

Answer:

D

Step-by-step explanation:

I believe D is the answer because 10^-3 = 0.001 ; there is only one answer that has 0.001 in the correct exponential form (10^-3).

Also, for the future, to correctly figure out what the exponential form, try to take the number given, let's say 10^-2 (^ means to the power of), and move the decimal place to the left, two decimal places (because of the power of -2).

Ex: 10.0 --> 0.01

Hope this helped!

3 0

3 years ago

What is the 14th term in the sequence an=-4n+13

mojhsa [17]

<span>We have the sequence that follows the pattern: a n = - 4 n + 13. If we want to calcelate the 14th term in this sequence ( or a 14 ), we have to substitute in this formula: n = 14. Then we will get: a 14 = - 4 * 14 + 13 a 14 = - 56 + 13, and finally: a 14 = - 43. Answer: The 14th term in the sequence is - 43.</span>

7 0

3 years ago

Identify the zeros of the polynomial function .

Komok [63]

Step-by-step explanation:

To start, every term has an

x

in common, so we can factor that out. Doing this, we get

x

(

x

2

+

x

−

42

)

=

0

What I have in blue can be factored by a thought experiment. What two numbers have a sum of

1

and a product of

−

42

?

After some trial and error, we arrive at

−

6

and

7

. Thus, this business can be factored as

x

(

x

−

6

)

(

x

+

7

)

=

0

We can leverage the zero product property here, setting all three terms equal to zero. As our zeroes, we get

x

=

0

,

x

=

6

, and

x

=

−

7

4 0

3 years ago

Can someone help me with this please

kiruha [24]

Answer:

Step-by-step explanation:

Answer is 30

3 0

3 years ago

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 2 cos(t) + sin(2t),[0, π/2].

castortr0y [4]

Answer:

The absolute maximum is $\frac{3\sqrt 3}2$ and the absolute minimum value is $0.$

Step-by-step explanation:

Differentiate of $f$ both sides w.r.t. $t$ ,

$f(t)=2 \cos t+\sin 2t$

$\Rightarrow f'(t)=-2\sin t+2\cos 2t$

Now take $f'(t)=0$

$\Rightarrow -2\sin t+2\cos 2t=0$

$\Rightarrow 2\cos 2t=2\sin t$

$\Rightarrow \cos 2t=\sin t$

$\Rightarrow 1-2\sin ^2t =\sin t \quad \quad [\because \cos 2t = 1-2\sin ^2t]$

$\Rightarrow 2\sin ^2t+\sin t-1=0$

$\Rightarrow 2\sin ^2t+2\sin t-\sin t-1=0$

$\Rightarrow 2\sin t(\sin t+1)-1(\sin t+1)=0$

$\Rightarrow (\sin t+1)(2\sin t-1)=0$

$\Rightarrow \sin t+1=0 \;\text{and}\; 2\sin t-1=0$

$\Rightarrow \sin t =-1 \;\text{and}\; \sin t =\frac 12$

In the interval $0\leq t\leq \frac {\pi}2$ , the answer to this problem is $\frac {\pi}6$

Now find the second derivative of $f(t)$ w.r.t. $t$ ,

$f''(t)=-2\cos t-4\sin 2t$

$\Rightarrow \left[f''(t)\right]_{t=\frac {\pi}6}=-2\times \frac {\sqrt 3}2-4\times \frac{\sqrt 3}2=-3\sqrt 3$

Thus, $f(t)$ is maximum at $t=\frac {\pi}6$ and minimum at $t=0$

$\left[f(t)\right]_{t=\frac {\pi}6}=2\times \frac {\sqrt 3}2+\frac{\sqrt 3}2=\frac{3\sqrt 3}2\;\text{and}\;\left[f(t)\right]_{t=\frac{\pi}2}= 2\times 0+0=0$

Hence, the absolute maximum is $\frac{3\sqrt 3}2$ and the absolute minimum value is $0$ .

7 0

3 years ago