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mojhsa [17]
4 years ago
13

Find the distance between points M(-1,-10) and (-12,-3). round to the nearest tenth

Mathematics
1 answer:
Kitty [74]4 years ago
4 0

The formula of the distance between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

We have the points (-1, -10) and (-12, -3). Substitute:

d=\sqrt{(-12-(-1))^2+(-3-(-10))^2}=\sqrt{(-11)^2+7^2}=\sqrt{121+49}=\sqrt{170}

\sqrt{170}\approx13.04

Answer: d ≈ 13.0

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1/2(y+3)>1/3(4-y) with the steps
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y

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3 0
3 years ago
A stick has a length of $5$ units. The stick is then broken at two points, chosen at random. What is the probability that all th
Zinaida [17]

Answer:

4/25 = 0.16

Step-by-step explanation:

The shortest stick must be between 0 and 5/3.  The probability that it is longer than 1 is therefore:

(5/3 − 1) / (5/3 − 0)

(2/3) / (5/3)

2/5

So the probability that both of the shortest sticks are longer than 1 is (2/5)² = 4/25.

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3 years ago
If the relationship is proportional, what is the missing value from the table?
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Please solve these 2 ;-;<br><br> y = 2x – 6<br> y = –4x + 3
Yuri [45]

Answer:

x = 3/2 | y = -3

Step-by-step explanation:

Given equations:

  • y = 2x - 6. . . .(i)
  • y = -4x + 3. . . . (ii)

Substituting from equation (i) for y:

==> 2x - 6 = -4x + 3

==> 2x + 4x = 6 + 3

==> 6x = 9

<em>dividing both </em><em>sides by 3</em><em>:</em>

==> 2x = 3

==> x = 3/2

Substituting 3/2 for x in equation (i):

==> y = 2(3/2) - 6

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3 0
3 years ago
Read 2 more answers
Given f(x) = (lnx)^3 find the line tangent to f at x = 3
kirill [66]
Explanation

We must the tangent line at x = 3 of the function:

f(x)=(\ln x)^3.

The tangent line is given by:

y=m*(x-h)+k.

Where:

• m is the slope of the tangent line of f(x) at x = h,

,

• k = f(h) is the value of the function at x = h.

In this case, we have h = 3.

1) First, we compute the derivative of f(x):

f^{\prime}(x)=\frac{d}{dx}((\ln x)^3)=3*(\ln x)^2*\frac{d}{dx}(\ln x)=3*(\ln x)^2*\frac{1}{x}=\frac{3(\ln x)^2}{x}.

2) By evaluating the result of f'(x) at x = h = 3, we get:

m=f^{\prime}(3)=\frac{3}{3}*(\ln3)^2=(\ln3)^2.

3) The value of k is:

k=f(3)=(\ln3)^3

4) Replacing the values of m, h and k in the general equation of the tangent line, we get:

y=(\ln3)^2*(x-3)+(\ln3)^3.

Plotting the function f(x) and the tangent line we verify that our result is correct:

Answer

The equation of the tangent line to f(x) and x = 3 is:

y=(\ln3)^2*(x-3)+(\ln3)^3

6 0
1 year ago
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