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sergeinik [125]

3 years ago

8

Adrian sells handmade puppets for $10 apiece. One puppet gets damaged, so Adrian sells it for $6, resulting in a loss of $4. He

asks his assistant to enter the loss in the account book. What should the entry be? Type your answer as a numeral.

2 answers:

Romashka [77]3 years ago

5 0

It would be a loss of 4 dollars.

Pachacha [2.7K]3 years ago

5 0

-4 i think, because they lost 4 dollars but im not sure

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The radius of a sphere decreases at a rate of 6 m/sec. find the rate at which the surface area decreases when the radius is 20 m

DanielleElmas [232]

The radius of the sphere decreases at the rate of 6 m/ sec

That is, $\frac{dr}{dt}=-6$

Surface Area of the sphere of radius r is given by $S=4\pi r^2$

Differentiating with respect to t,

$\frac{dS}{dt}=4\pi (2r)\frac{dr}{dt}=8\pi r\frac{dr}{dt}$

$\frac{dS}{dt}=8\pi (20)^2(-6)=-960\pi \: m^2/sec$

8 0

3 years ago

a line passes through the point (6, -9) and has a slope of 3/2. write an equation in point-slope form for this line

Stella [2.4K]

Y=3/2x is the answer

3 0

2 years ago

Read 2 more answers

the area of a rectangle is 90 in2^. the ratio of the length to the width is 5:2. find the length and the width

-BARSIC- [3]

Length and width of rectangle is 15 inches and 6 inches respectively

<h3><u>Solution:</u></h3>

Given that area of a rectangle is 90 square inch

Ratio of length to the width = 5: 2.

Need to determine length and width of rectangle.

As ratio of length to the width is 5 : 2

Lets assume length of rectangle = 5x inches and width of rectangle = 2x inches.

<em><u>The formula for area of rectangle is given as:</u></em>

$\text { Area of rectangle }=\text { length of rectangle } \times \text { width of rectangle}$

Substituting the given value of area of rectangle and assumed value of length and width of rectangle we get:

$\begin{array}{l}{90=5 x \times 2 x} \\\\ {=>90=10 x^{2}}\end{array}$

On solving the above expression for x we get

$\begin{array}{l}{=>\frac{90}{10}=x^{2}} \\\\ {=>x^{2}=9} \\\\ {=>x=\sqrt{9}=3}\end{array}$

$\begin{array}{l}{\text { Length of rectangle }=5 \times x=5 \times 3=15 \text { inches }} \\\\ {\text { Width of rectangle }=2 \times} x=2 \times} 3=6 \text { inches }}\end{array}$

Hence length and width of rectangle is 15 inches and 6 inches.

4 0

3 years ago

Item 10 Find the amount of time. I=$450, P=$2400, r=7.5%

Alex787 [66]

The answer is:  "2.5 years" .
___________________________________________________
Note: I = P * r * t ; { " Interest = Principal * rate * time "} ;

→ Solve for "t" {"time", in years} ;

Divide each side of the equation by "{P * r}" ;
to isolate "t" on one side of the equation ;

→ I / (P * r) = {P * r * t) / (P * r} ;

to get: " I / (P * r) = t " ;

  ↔ t = I / (P * r) ;

Given: I = $450 ;

  <span>P = $2400 ;

r = 7.5% = 7.5/100 = 0.075 ;

Plug in these values into the formula to solve for the time, "t" :

  </span>→ t = I / (P * r ) ;

= $450 / (<span>$2400 * 0.075) ;

= </span>$450 / ($2400 * 0.075) ;

= $450 / $180 ;

= $45 / $18 ;

= ($45 ÷ 9) / ($18 ÷ 9)

= $5 / $2 ;

= 2.5 ;

→ t = 2.5 years.
_______________________________________________________
The answer is: "2.5 years" .
_______________________________________________________

7 0

3 years ago

Answer the question in the picture

nekit [7.7K]

Recall the angle sum identities:

$\sin(x+y)=\sin x\cos y+\cos x\sin y$

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

Now,

$\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$

Divide through numerator and denominator by $\cos x\cos y$ to get

$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$

Next, we use the fact that $x,y$ lie in the first quadrant to determine that

$\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2$

$\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}$

So we then have

$\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

$\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1$

Finally,

$\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73$

4 0

3 years ago