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2 years ago

HELPPP i need answers asappppp

Mathematics

1 answer:

sweet-ann [11.9K]2 years ago

6 0

Answer:

The expressions are equivalent

Step-by-step explanation:

We can start by simplifying the first expression:

1.25x+4+0.75x-3

Move like terms so that they are together:

1.25x+0.75x+4-3

Now, combine like terms:

2x+1

The expressions are equivalent

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A rope of length 18 feet is arranged in the shape of a sector of a circle with central angle O radians, as shown in the

creativ13 [48]

Answer:

$A(\theta)=\frac{162 \theta}{(\theta+2)^2}$

Step-by-step explanation:

The picture of the question in the attached figure

step 1

Let

r ---> the radius of the sector

s ---> the arc length of sector

Find the radius r

we know that

$2r+s=18$

$s=r \theta$

$2r+r \theta=18$

solve for r

$r=\frac{18}{2+\theta}$

step 2

Find the value of s

$s=r \theta$

substitute the value of r

$s=\frac{18}{2+\theta}\theta$

step 3

we know that

The area of complete circle is equal to

$A=\pi r^{2}$

The complete circle subtends a central angle of 2π radians

using proportion find the area of the sector by a central angle of angle theta

Let

A ---> the area of sector with central angle theta

$\frac{\pi r^{2} }{2\pi}=\frac{A}{\theta} \\\\A=\frac{r^2\theta}{2}$

substitute the value of r

$A=\frac{(\frac{18}{2+\theta})^2\theta}{2}$

$A=\frac{162 \theta}{(\theta+2)^2}$

Convert to function notation

$A(\theta)=\frac{162 \theta}{(\theta+2)^2}$

6 0

3 years ago

PLEASE HELP! INVERSE QUESTION!

Katena32 [7]

Last one.

Hope this helps.

3 0

3 years ago

A used-car dealer sold one car at a profit of 25 percent of the dealer's purchase price for that car and sold another car at a l

Daniel [21]

Answer:

C. 1000 loss.

Step-by-step explanation:

Let x represent purchase price of each car.

We have been given that a used-car dealer sold one car at a profit of 25 percent of the dealer's purchase price for that car and sold another car at a loss of 20 percent of the dealer's purchase price for that car.

We can represent the car that dealer sold with 25% profit to find purchase price as:

$x+0.25x=20,000$

$1.25x=20,000$

$\frac{1.25x}{1.25}=\frac{20,000}{1.25}$

$x=16,000$

Therefore, the purchase price of 1st car was $16,000.

We can represent the car that dealer sold with 20% loss to find purchase price as:

$x-0.20x=20,000$

$0.80x=20,000$

$\frac{0.80x}{0.80}=\frac{20,000}{0.80}$

$x=25,000$

Therefore, the purchase price of 2nd car was $25,000.

The total purchase price of both cars would be $16,000+25,000=41,000$

The total sale price of both cars $20,000+20,000=40,000$ .

We can see that the sale price of both car is less than purchase price by $1000, so the dealer got a loss of $1000.

Therefore, the dealer's total loss, in dollars, for the two transactions combined was 1000 and option C is the correct choice.

4 0

3 years ago

10.04 × 8.8= ?<br><br>can you help me again please

ANEK [815]

Answer:

10.04 × 8.8 =?, ? = 88.352 :)

7 0

3 years ago

Read 2 more answers

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,

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

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

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

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,

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

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