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Reil [10]
3 years ago
12

Find the graph of y=-3/4X-2

Mathematics
1 answer:
morpeh [17]3 years ago
3 0
To graph the line y=- \frac{3}{4}x-2 we are going to take advantage of the fact that its y-intercept is -2 when x=0, so our first point is (0,-2). To find our second point, we are going to find the x-intercept; to do it, we are going to set y=0 and solve for x:
y=- \frac{3}{4}x-2
0=- \frac{3}{4}x-2
2=- \frac{3}{4}x
x= \frac{2}{- \frac{3}{4} }
x=- \frac{8}{3}
Now that we have our second point (0,- \frac{8}{3} ), we just need to join our two points with a straight line.

We can conclude that the graph of <span>y=-3/4X-2 is:</span>

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What is the area of the shaded region?​
IRINA_888 [86]
20 because of the whole is 40 and the shaded part is by 20 you just take half of 40 and you get 20
3 0
3 years ago
1/3 (6x +12) - 2(x-7)=19
Morgarella [4.7K]

Answer:

no solution

Step-by-step explanation:

Given

\frac{1}{3}(6x + 12) - 2(x - 7) = 19 ← distribute parenthesis on left side

2x + 4 - 2x + 14 = 19 , that is

18 = 19 ← not possible

This indicates the equation has no solution

4 0
3 years ago
The expression 62.4d-21062.4d−21062, point, 4, d, minus, 210 gives the number of Indian rupees you receive when you exchange \$d
Arisa [49]
<span>601.2 rupees You've been given the equation: r = 62.4d - 210 I will assume that 62.4 is the actual exchange rate and that the 210 is a service fee of some sort for performing the exchange. So with 13 dollars, let's substitute that value into the equation and calculate: r = 62.4d - 210 r = 62.4*13 - 210 r = 811.2 - 210 r = 601.2 So you'll get 601.2 rupees.</span>
8 0
3 years ago
Read 2 more answers
Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to f
tino4ka555 [31]

Answer:

The minimum value of the given function is f(0) = 0

Step-by-step explanation:

Explanation:-

Extreme value :-  f(a, b) is said to be an extreme value of given function 'f' , if it is a maximum or minimum value.

i) the necessary and sufficient condition for f(x)  to have a maximum or minimum at given point.

ii)  find first derivative f^{l} (x) and equating zero

iii) solve and find 'x' values

iv) Find second derivative f^{ll}(x) >0 then find the minimum value at x=a

v) Find second derivative f^{ll}(x) then find the maximum value at x=a

Problem:-

Given function is f(x) = log ( x^2 +1)

<u>step1:</u>- find first derivative f^{l} (x) and equating zero

  f^{l}(x) = \frac{1}{x^2+1} \frac{d}{dx}(x^2+1)

f^{l}(x) = \frac{1}{x^2+1} (2x)  ……………(1)

f^{l}(x) = \frac{1}{x^2+1} (2x)=0

the point is x=0

<u>step2:-</u>

Again differentiating with respective to 'x', we get

f^{ll}(x)=\frac{x^2+1(2)-2x(2x)}{(x^2+1)^2}

on simplification , we get

f^{ll}(x) = \frac{-2x^2+2}{(x^2+1)^2}

put x= 0 we get f^{ll}(0) = \frac{2}{(1)^2}   > 0

f^{ll}(x) >0 then find the minimum value at x=0

<u>Final answer</u>:-

The minimum value of the given function is f(0) = 0

5 0
3 years ago
(1.3 x 10-2) (0.05) = ?
Alik [6]

Answer:

=0.065x10−0.1

Step-by-step explanation:

(1.3x10−2)(0.05)

=(1.3x10+−2)(0.05)

=(1.3x10)(0.05)+(−2)(0.05)

=0.065x10−0.1

7 0
3 years ago
Read 2 more answers
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