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3 years ago

What’s 17/5 as a decimal?

Mathematics

2 answers:

DanielleElmas [232]3 years ago

8 0

It’s 3.4 because all your doing is dividing 17 into 5

nikdorinn [45]3 years ago

6 0

Answer:

The answer is 3.4

Hope this helps

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One ordered pair (a,b) satisfies the two equations ab^4 = 384 and a^2 b^5 = 4608. What is the value of a in this ordered pair?

Alekssandra [29.7K]

Answer: a = 3∛2

<u>Step-by-step explanation:</u>

ab⁴ = 384 --> a = 384/b⁴

Substitute a = 384/b⁴ into the second equation to solve for "b".

a²b⁵ = 4608

$\bigg(\dfrac{384}{b^4}\bigg)^2\cdot b^5=4608\\\\\\\dfrac{147,456b^5}{b^8}=4608\\\\\\\dfrac{147,456}{b^3}=4608\\\\\\\dfrac{147,456}{4608}=b^3\\\\\\32=b^3\\\\\\\sqrt[3]{32} =b\\\\\\2\sqrt[3]{4} =b$

Substitute b = 2∛4 into the first equation to solve for "a".

ab⁴ = 384

a(2∛4)⁴ = 384

a = 384/(2∛4)⁴

a = 24/4∛4

= 6/∛4

= 6(∛2)/2

= 3∛2

5 0

2 years ago

What is 25% as a decimal

Artist 52 [7]

25% as a fraction is 1/4 or 25/100 so as a decimal its 1/4 of 1 which is .25

3 0

3 years ago

Read 2 more answers

F(x) = -x² – 10x + 16<br> Find f(-7)

Gala2k [10]

Answer:

-5

Step-by-step explanation:

or, f(x) = -x^2+10x+16

or, f(7) = (-7)^2+10*-7+16 (-7*-7=49)

=49-70+16

=65-70

=-5

5 0

2 years ago

Sue made a batch of snack mix for the party. She combined 12 ounces of mixed nuts, 20 ounces of cereal, and 8 ounces of pretzels

Levart [38]

Answer:

The pretzels are 20% of the total weight of the snack.

Step-by-step explanation:

Total weight 12 + 20 + 8 = 40

So 8 / 40 = 20%

6 0

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

2 years ago