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3 years ago

Solve the system of linear equations by substitution.

Mathematics

1 answer:

Bezzdna [24]3 years ago

3 0

For this you need to solve by elmination

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Solve for f.<br><br> 8 = 2f + 2

tiny-mole [99]

Answer:

f = 3

Step-by-step explanation:

8 = 2f + 2

-2 -2

6 = 2f

/2 /2

3 = f

5 0

3 years ago

What is the value of b in the equation 3b + 2(b - 1) =8

WARRIOR [948]

Answer:

b = 2

Step-by-step explanation:

In order to find the value of b, we have to isolate b on one side of equation first

So,

$3b+2(b-1)=8$

Multiplying 2 with b-1

$3b+2b-2=8\\5b-2=8$

Adding 2 on both sides

$5b-2+2=8+2\\5b=10$

Dividing by 5 on both sides

$\frac{5b}{5}=\frac{10}{5} \\b=2$

So the value of b is 2 ..

6 0

3 years ago

Read 2 more answers

An assistant at an animal shelter wants to know the typical age of dogs admitted to the shelter. the shelter has admitted 353 do

Nikitich [7]

Step-by-step explanation:

hope you can understand

5 0

2 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

A strawberry weighs about 1 out of ten an orange. if an orange weighs about 67 grams, about how many grams does a strawberry wei

andre [41]

Answer:

16.75 g

Step-by-step explanation:

Orange's weight = 67 g

Strawberry's weight

= ¼ of the weight of orange

= ¼ × weight of orange

= ¼ × 67 g

= 67 g/4

= 16.75 g

8 0

3 years ago