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PSYCHO15rus [73]

2 years ago

6

Update for the first one

2 answers:

Zepler [3.9K]2 years ago

8 0

Answer is x+2.5 Hope it helped

PSYCHO15rus [73]2 years ago

7 0

Answer:

y = x + 2.5

Step-by-step explanation:

slope = (y2-y1)/(x2-x1) where the variables indicate the coordinates of the two points

slope = (2.5-1.5)/(0+1) = 1

equation = y-y1 = m(x-x1)

y - 2.5 = x

y = x + 2.5

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b). Perimeter = 33.7 units

Step-by-step explanation:

Vertices of the triangle ABC are A(-4, -2), B(1, 7) and C(8, -2).

(a). Area of the triangle ABC = $\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$ (Absolute value)

By substituting the values from the given vertices,

Area = $\frac{1}{2}[(-4)(7+2)+(1)(-2+2)+8(-2-7)]$

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(b). Distance between two vertices (a, b) and (c, d)

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AB = $\sqrt{(-4-1)^{2}+(-2-7)^{2}}$

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= 11.402 units

AC = $\sqrt{(-4-8)^2+(-2+2)^2}$

= 12 units

Perimeter of the triangle = AB + BC + AC = 10.295 + 11.402 + 12

= 33.697

≈ 33.7 units

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Find the error & find the correct answer 2In(x)=In(3x)-[In(9)-2In(3)] In(x^2)=In(3x)-[In(9)-In(9)] In(x^2)=In(3x)-0 In(x^2)=

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Answer:

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Solution:x=0 and 3

Step-by-step explanation:

We have to find the error and correct answer

Given: $2ln x=ln(3x)-[ln9-2ln(3)]$

$lnx^2=ln(3x)-[ln9-ln3^2]$

Using the formula

$alog b=logb^a$

$lnx^2=ln(3x)-[ln9-ln9]$

$lnx^2=ln(3x)-0$

$lnx^2=ln(3x)$

$x^2=3x$

$x^2-3x=0$

$x(x-3)=0$

Therefore, x=0 and x=3

But last step in the given solution

$lnx^2=ln\frac{3x}{0}=\infty$

It is wrong this property is used when

$log m-log n$ then

$log\frac{m}{n}$

Hence, the student wrote $lnx^2=ln\frac{3x}{0}$ instead of $lnx^2=ln3x$ and solution is given by

x=0 and x=3

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2 years ago