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2 years ago

Pls someone help me do this will give brainless

Mathematics

1 answer:

kenny6666 [7]2 years ago

3 0

Answer:

Step-by-step explanation:

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Complete the solution of the equation. Find the value of y when x equals 15.

Mariulka [41]

Answer:

y = -6 PLEASE GIVE BRAINLIEST

Step-by-step explanation:

-15 - 9y = 39

39 + 15 = 54

-9y =54

y = 54 ÷ -9 = -6

y = -6

8 0

2 years ago

HELP!! Not hard, just confusing.. ΔABC is reflected across the x-axis and then translated 4 units up to create ΔA′B′C′. What are

allsm [11]

The coordinates of the vertices of ΔABC are:A( x1, y1), B( x2, y2) and C( x3, y 3 ). After it is reflected across the x-axis, coordinates are ( x1, -y1), (x2, -y2), (x3, -y3). Finally, the coordinates of the vertices of ΔA´B´C´ after translation are: A´( x1, 4-y1), B´( x2, 4- y2), C´( x3, 4-y3 )

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2 years ago

60 POINTS + BRAINLIEST

Nitella [24]

Answer:

Step-by-step explanation:

By tracing the figure, we see that we go around a semicricle twice and go in a straight path twice. So the perimeter of this figure consists of two semicircles and two line segments, hence the answer is C.

3 0

2 years ago

Read 2 more answers

Please answer this correctly without making mistakes I want ace expert and genius people to answer this correctly without making

marin [14]

Answer:

24 i think

Step-by-step explanation:

I evaluated using he given value

4 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

2 years ago