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forsale [732]
3 years ago
10

Identify the domain of the exponential function shown in the following graph: WARNING ITS NOT D

Mathematics
1 answer:
mamaluj [8]3 years ago
4 0
Answer is B.
The x-values of this function are all real numbers greater than or equal to zero.
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The triangle shown below has an area of 22 units squared. The height is 4 units. Find the base of the triangle.
Papessa [141]

Answer:

11 units

Step-by-step explanation:

the formula is (1/2)bh

here we have(1/2)b(4)

the area is 22, so (1/2)b(4)=22

plug in 11 and its correct

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At the start of an experiment, the temperature of a solution was -12°C. During the expirament
Rom4ik [11]

Answer:

The final temperature of the solution was 5°C

Step-by-step explanation:

we know that

1) At the start of an experiment, the temperature of a solution was -12°C

In this moment the temperature is -12°C

2) During the experiment, the temperature of the solution rose 5°C each hour during 4 hours

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3 years ago
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Definition: An equation that involves a square root. One or both sides of the equation may contain the square root.
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4 years ago
Write your question here (Keep it simple and clear to get the best answer)striangle has vertices A(2;5); B(1;-2) and C(-5;1). De
katovenus [111]
<h2>Answer:</h2>

(a) y = \frac{-1}{2}x -  \frac{3}{2}

(b) y = 2x + 3

<h2>Step-by-step explanation:</h2>

(a) The equation of a line given by points M(x₁, y₁) and N(x₂, y₂) is given by:

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

Where;

m = \frac{y_2 - y_1}{x_2 - x_1} = slope or gradient of the line  ---------------(ii)

Given points on the triangle are:

A(2,5)

B(1,-2)

C(-5,1)

To find the equation of line BC, we use the formulas in equations (i) and (ii) where the points of the line are B(1,-2) and C(-5,1) and;

x₁ = 1

y₁ = -2

x₂ = -5

y₂ = 1

==> First get the gradient using equation (ii) as follows;

m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-2)}{-5 -1}

m =   \frac{3}{-6}

m = \frac{-1}{2}

==> Now, use equation (i) to find the equation of the line as follows;

y - (-2) = \frac{-1}{2}(x - 1)

y + 2 = \frac{-1}{2}(x - 1)

<em>Multiply both sides by 2</em>

2(y+2) = -1 ( x - 1 )

2y + 4 = -x + 1

2y = -x + 1 - 4

2y = - x - 3

y =  \frac{-1}{2}x -  \frac{3}{2}

Therefore, the equation of the line is y = \frac{-1}{2}x -  \frac{3}{2}

(b) To find the perpendicular line from A to BC, note that

i. two lines are perpendicular if they meet at 90°

ii. the general equation of a line could also be written as <em>y = mx + c</em> where m is the slope and c is the intercept.

iii. when one line has a slope of m, then a perpendicular line to that line will have a slope of \frac{-1}{m}

The equation of line BC is y = \frac{-1}{2}x -  \frac{3}{2}.

This means that BC has a slope of \frac{-1}{2}

A perpendicular line from A to BC will have a slope of 2.

Now to get the equation of this perpendicular line from A(2, 5) to BC, we use the general equation of a line given in equation (i)

where;

m = 2

x₁ = 2

y₁ = 5

Substitute these values into equation (i)

y - 5 = 2(x - 2)

Solving by simplification gives;

y - 5 = 2x - 2

y = 2x + 3

Therefore, the equation of the perpendicular line is y = 2x + 3

5 0
3 years ago
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