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3 years ago

A set of 12 windup toys costs $9. What is the cost per windup toy?

Mathematics

2 answers:

-BARSIC- [3]3 years ago

4 0

Answer:

0.75

Step-by-step explanation:

9 ÷ 12 = 0.75

hope this helps...

Ann [662]3 years ago

4 0

Answer:

9/12 = 3/4 = 0.75

Step-by-step explanation:

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A tower that is 111 feet tall casts a shadow 126 feet long. find the angle of elevation of the sun to the nearest degrre

Sergeeva-Olga [200]

The angle here is an acute angle at ground level. Then the side adjacent to this angle is 126 feet. The opp. side is the tower, 111 feet tall.

Then

tan theta = opp / adj = 111 / 126. Using the inverse tangent function, we find that the measure of the angle is 0.722 radian or 41 degrees (to the nearest degree).

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Brian was thinking of a number. Brian halves the number and gets an answer of 67.8. Form an equation with x from the information

Fofino [41]

Answer:

x ÷ 2 = 67.8

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3 years ago

Can someone please help me with numbers 1, a, b, c, 2, a, b, c

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3 years ago

A triangular frame has sides that measure 15'-7",20'-4" and 26'-2". What is the total length of three sides?

tatuchka [14]

we know that

$1\ foot=12\ inches$

$1'=12"$

we know that

side 1=15'-7"

side 2=20'-4"

side 3=26'-2"

To find the total length of three sides sum the three sides

total length=side 1+side 2 +side 3

substitute

total length=15'-7"+20'-4"+26'-2"

total length=(15'+20'+26')+(7"+4"+2")

total length=(61')+(13")

remember that

12"=1'

13"=12"+1"=1'+1"

substitute

total length=(61')+(1'+1")

total length=(62')+(1")---------> 62'-1"

therefore

<u>the answer is</u>

the total length of three sides is 62'-1"

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3 years ago

Use the Newton-Raphson method to find the root of the equation f(x) = In(3x) + 5x2, using an initial guess of x = 0.5 and a stop

xxMikexx [17]

Answer with explanation:

The equation which we have to solve by Newton-Raphson Method is,

f(x)=log (3 x) +5 x²

$f'(x)=\frac{1}{3x}+10 x$

Initial Guess =0.5

Formula to find Iteration by Newton-Raphson method

$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\\\\x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}\\\\ x_{1}=0.5-\frac{\log(1.5)+1.25}{\frac{1}{1.5}+10 \times 0.5}\\\\x_{1}=0.5- \frac{0.1760+1.25}{0.67+5}\\\\x_{1}=0.5-\frac{1.426}{5.67}\\\\x_{1}=0.5-0.25149\\\\x_{1}=0.248$

$x_{2}=0.248-\frac{\log(0.744)+0.30752}{\frac{1}{0.744}+10 \times 0.248}\\\\x_{2}=0.248- \frac{-0.128+0.30752}{1.35+2.48}\\\\x_{2}=0.248-\frac{0.17952}{3.83}\\\\x_{2}=0.248-0.0468\\\\x_{2}=0.2012$

$x_{3}=0.2012-\frac{\log(0.6036)+0.2024072}{\frac{1}{0.6036}+10 \times 0.2012}\\\\x_{3}=0.2012- \frac{-0.2192+0.2025}{1.6567+2.012}\\\\x_{3}=0.2012-\frac{-0.0167}{3.6687}\\\\x_{3}=0.2012+0.0045\\\\x_{3}=0.2057$

$x_{4}=0.2057-\frac{\log(0.6171)+0.21156}{\frac{1}{0.6171}+10 \times 0.2057}\\\\x_{4}=0.2057- \frac{-0.2096+0.21156}{1.6204+2.057}\\\\x_{4}=0.2057-\frac{0.0019}{3.6774}\\\\x_{4}=0.2057-0.0005\\\\x_{4}=0.2052$

So, root of the equation =0.205 (Approx)

Approximate relative error

$=\frac{\text{Actual value}}{\text{Given Value}}\\\\=\frac{0.205}{0.5}\\\\=0.41$

Approximate relative error in terms of Percentage

=0.41 × 100

= 41 %

7 0

3 years ago