Hellpppppppppp pleaseee

what is the decimal equivalent of the fraction? 7/12 enter your answer to the nearest thousandths in the box

Likurg_2 [28]

The answer is 0.583, and the 3 has a bar notation

Hope this helped

btw i also do k12

4 0

2 years ago

If Cody is draining his hot tub which holds 1,400 gallons of water at a rate of 27 gallons a minute, what impact will it have on

inessss [21]

Facts:

- tub has 1400 gallons

- draining at 27 gallons/minute

How much water will be left after 18 minutes?

We know that after 1 minute, there will be 27 gallons less.

So after 18 minutes, there must be 18 times less water.

27 x 18 = 486

This is how much water was lost in 18 min

So to find out how much he had left after 486 is subtracted.

Therefore, the answer is A) -486 gallons

4 0

3 years ago

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

b)

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

c)

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$

And,

$f'(A)=\frac{dB}{dA}=\sqrt{3}$ from part b

Therefore, the linear approximation at $A=\frac{\pi}{4}$ is,

$f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}$

d)

Use part (c), when $A=46°$ , B is approximately,

$B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°$

8 0

3 years ago

Scott and Harry went cycling every day. They increased the number of minutes of cycling every week as described below:

djyliett [7]

Answer:

(A)

Step-by-step explanation:

Scott cycled for 10 minutes in first week, 15 minutes in second week,20 minutes in third week and 25 minutes in the fourth week which maintains a consistency in the line as there is consistent slope when drawn on a graph.

On, the other hand, harry cycled for 10 minutes in first week, 20 minutes in second, 40 minutes in third and 80 minutes in fourth week in which there is no consistency in the timings according to the weeks. Therefore, only, scott's method is linear as the number of minutes increased by an equal factor every week.

4 0

3 years ago

Read 2 more answers

-5/8 ÷ 9/10 A. -25/36 B. -9/16 C. -45/80 D. -1/36

Oduvanchick [21]

A. -25/36

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5 0

2 years ago

Read 2 more answers

Hellpppppppppp pleaseee​

Hellpppppppppp pleaseee