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3 years ago

What is x in the equation-5/8 x=-160

Mathematics

2 answers:

Rudik [331]3 years ago

7 0

Answer:256

Step-by-step explanation:

-160×8/-5=x

X=256

marusya05 [52]3 years ago

4 0

Answer:

x=256

Step-by-step explanation:

-5/8x=-160- divide both sides by the reciprocal of the coefficient of x

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Answer:I think the answer is c

Step-by-step explanation:

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2 years ago

What is the value of -9 + (-4 + 7) (2)?

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Step-by-step explanation:

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2 years ago

Read 2 more answers

A triangle is graphed in the coordinate plane. The vertices of the triangle have coordinates (–3, 1), (1, 1), and (1, –2). What

vladimir2022 [97]

Answer:

The perimeter of the triangle is $12\ units$

Step-by-step explanation:

Let

$A(-3,1),B(1,1),C(1,-2)$

we know that

The perimeter of triangle is equal to

$P=AB+BC+AC$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

$A(-3,1),B(1,1)$

substitute in the formula

$AB=\sqrt{(1-1)^{2}+(1+3)^{2}}$

$AB=\sqrt{(0)^{2}+(4)^{2}}$

$AB=4\ units$

step 2

Find the distance BC

$B(1,1),C(1,-2)$

substitute in the formula

$BC=\sqrt{(-2-1)^{2}+(1-1)^{2}}$

$BC=\sqrt{(-3)^{2}+(0)^{2}}$

$BC=3\ units$

step 3

Find the distance AC

$A(-3,1),C(1,-2)$

substitute in the formula

$AC=\sqrt{(-2-1)^{2}+(1+3)^{2}}$

$AC=\sqrt{(-3)^{2}+(4)^{2}}$

$AC=5\ units$

step 4

Find the perimeter

$P=AB+BC+AC$

substitute the values

$P=4+3+5=12\ units$

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

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

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

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

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