What is another way to write 7 8/10

In a complete paragraph, explain how to scope a speech. <br> Please use your own words.

butalik [34]

Answer:

Prepare. You dread public speaking because you don't know what to say.

Psych yourself. ...

Know your audience. ...

Be an expert on your topic. ...

Visualize. ...

Conduct an ocular inspection. ...

Read, watch, and listen to good speakers.

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What do you do to the equation y = x to make its graph move up on the y-axis?

densk [106]

Recall that in Linear Functions, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 8. We can see right away that the graph crosses the y-axis at the point (0, 4) so this is the y-intercept.

Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point (–2, 0). To get from this point to the y-intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be

\displaystyle m=\frac{\text{rise}}{\text{run}}=\frac{4}{2}=2m=

run

rise

=

2

4

=2

Substituting the slope and y-intercept into the slope-intercept form of a line gives

\displaystyle y=2x+4y=2x+4

HOW TO: GIVEN A GRAPH OF LINEAR FUNCTION, FIND THE EQUATION TO DESCRIBE THE FUNCTION.

Identify the y-intercept of an equation.

Choose two points to determine the slope.

Substitute the y-intercept and slope into the slope-intercept form of a line.

EXAMPLE 4: MATCHING LINEAR FUNCTIONS TO THEIR GRAPHS

Match each equation of the linear functions with one of the lines in Figure 9.

\displaystyle f\left(x\right)=2x+3f(x)=2x+3

\displaystyle g\left(x\right)=2x - 3g(x)=2x−3

\displaystyle h\left(x\right)=-2x+3h(x)=−2x+3

\displaystyle j\left(x\right)=\frac{1}{2}x+3j(x)=

2

1

x+3

Graph of three lines, line 1) passes through (0,3) and (-2, -1), line 2) passes through (0,3) and (-6,0), line 3) passes through (0,-3) and (2,1)

Figure 9

SOLUTION

Analyze the information for each function.

This function has a slope of 2 and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function g has the same slope, but a different y-intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through (0, 3) so f must be represented by line I.

This function also has a slope of 2, but a y-intercept of –3. It must pass through the point (0, –3) and slant upward from left to right. It must be represented by line III.

This function has a slope of –2 and a y-intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right.

This function has a slope of \displaystyle \frac{1}{2}

2

1

and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. Lines I and II pass through (0, 3), but the slope of j is less than the slope of f so the line for j must be flatter. This function is represented by Line II.

Now we can re-label the lines as in Figure 10.

Figure 10

Finding the x-intercept of a Line

So far, we have been finding the y-intercepts of a function: the point at which the graph of the function crosses the y-axis. A function may also have an x-intercept, which is the x-coordinate of the point where the graph of the function crosses the x-axis. In other words, it is the input value when the output value is zero.

To find the x-intercept, set a function f(x) equal to zero and solve for the value of x. For example, consider the function shown.

\displaystyle f\left(x\right)=3x - 6f(x)=3x−6

Set the function equal to 0 and solve for x.

⎧

⎪

⎨

⎪

⎩

0

=

3

x

−

6

=

3

x

2

=

x

=

2

The graph of the function crosses the x-axis at the point (2, 0).

Q & A

Do all linear functions have x-intercepts?

No. However, linear functions of the form y = c, where c is a nonzero real number are the only examples of linear functions with no x-intercept. For example, y = 5 is a horizontal line 5 units above the x-axis. This function has no x-intercepts.

Graph of y = 5.

Figure 11

A GENERAL NOTE: X-INTERCEPT

The x-intercept of the function is value of x when f(x) = 0. It can be solved by the equation 0 = mx + b.

EXAMPLE 5: FINDING AN X-INTERCEPT

Find the x-intercept of \displaystyle f\left(x\right)=\frac{1}{2}x - 3f(x)=

2

1

x−3.

SOLUTION

Set the function equal to zero to solve for x.

\displaystyle \begin{cases}0=\frac{1}{2}x - 3\\ 3=\frac{1}{2}x\\ 6=x\\ x=6\end{cases}

⎩

⎪

⎨

⎪

⎧

0=

2

1

x−3

3=

2

1

x

6=x

x=6

The graph crosses the x-axis at the point (6, 0).

Analysis of the Solution

A graph of the function is shown in Figure 12. We can see that the x-intercept is (6, 0) as we expected.

Figure 12. The graph of the linear function \displaystyle f\left(x\right)=\frac{1}{2}x - 3f(x)=

2

1

5 0

3 years ago

A company rents bicycles to customers. The company charges an initial fee

photoshop1234 [79]

The hourly rate is $3 a hour with a $2 initial fee.

5 0

3 years ago

1. The height of a triangle is 6 m more than its base. The area of the triangle is 56 m². What is the length of the base? Enter

Elodia [21]

Answers:
1. 8 m
2. 17 m
3. 7 cm
4. 2 s

Explanations:

1. Let x = length of the base
x + 6 = height of the base

Then, the area of the triangle is given by

(Area) = (1/2)(base)(height)
56 = (1/2)(x)(x + 6)
56 = (1/2)(x²  + 6x)

Using the symmetric property of equations, we can interchange both sides of equations so that

(1/2)(x²  + 6x) = 56

Multiplying both sides by 2, we have

x² + 6x = 112

The right side should be 0. So, by subtracting both sides by 112, we have

x² + 6x - 112 = 112 - 112
x² + 6x - 112 = 0

By factoring, x² + 6x - 112 = (x - 8)(x + 14). So, the previous equation becomes

(x - 8)(x +14) = 0

So, either

x - 8 = 0 or x + 14 = 0

Thus, x = 8 or x = -14. However, since x represents the length of the base and the length is always positive, it cannot be negative. Hence, x = 8. Therefore, the length of the base is 8 cm.

2. Let x = length of increase in both length and width of the rectangular garden

Then,

14 + x = length of the new rectangular garden
12 + x = width of the new rectangular garden

So,

(Area of the new garden) = (length of the new garden)(width of the new garden)

255 = (14 + x)(12 + x) (1)

Note that

(14 + x)(12 + x) = (x + 14)(x + 12)
= x(x + 14) + 12(x + 14)
= x² + 14x + 12x + 168
= x² + 26x + 168

So, the equation (1) becomes

255 = x² + 26x + 168

By symmetric property of equations, we can interchange the side of the previous equation so that

x² + 26x + 168 = 255

To make the right side becomes 0, we subtract both sides by 255:

x² + 26x + 168 - 255 = 255 - 255
x² + 26x - 87 = 0

To solve the preceding equation, we use the quadratic formula.

First, we let

a = numerical coefficient of x² = 1

Note: if the numerical coefficient is hidden, it is automatically = 1.

b = numerical coefficient of x = 26
c = constant term = - 87

Then, using the quadratic formula

$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} = \frac{-26 \pm \sqrt{26^2 - 4(1)(-87)} }{2(1)} 
\newline x = \frac{-26 \pm \sqrt{1,024} }{2}
\newline
\newline x = \frac{-26 \pm 32 }{2}$

So,

$x = \frac{-26 + 32 }{2} \text{ or } x = \frac{-26 - 32 }{2}
\newline x = \frac{6 }{2} \text{ or } x = \frac{-58 }{2}
\newline \boxed{ x = 3 \text{ or } x = -29}$

Since x represents the amount of increase, x should be positive.

Hence x = 3.

Therefore, the length of the new garden is given by

14 + x = 14 + 3 = 17 m.

3. The area of the shaded region is given by

(Area of shaded region) = π(outer radius)² - π(inner radius)²
= π(2x)² - π6²
= π(4x² - 36)

Since the area of the shaded region is 160π square centimeters,

π(4x² - 36) = 160π

Dividing both sides by π, we have

4x² - 36 = 160

Note that this equation involves only x² and constants. In these types of equation we get rid of the constant term so that one side of the equation involves only x² so that we can solve the equation by getting the square root of both sides of the equation.

Adding both sides of the equation by 36, we have

4x² - 36 + 36 = 160 + 36
4x² = 196

Then, we divide both sides by 4 so that

x² = 49

Taking the square root of both sides, we have

$x = \pm 7$

Note: If we take the square root of both sides, we need to add the plus minus sign $(\pm)$ because equations involving x² always have 2 solutions.

So, x = 7 or x = -7.

But, x cannot be -7 because 2x represents the length of the outer radius and so x should be positive.

Hence x = 7 cm

4. At time t, h(t) represents the height of the object when it hits the ground. When the object hits the ground, its height is 0. So,

h(t) = 0 (1)

Moreover, since $v_0 = 27$ and $h_0 = 10$ ,

h(t) = -16t² + 27t + 10 (2)

Since the right side of the equations (1) and (2) are both equal to h(t), we can have

-16t² + 27t + 10 = 0

To solve this equation, we'll use the quadratic formula.

Note: If the right side of a quadratic equation is hard to factor into binomials, it is practical to solve the equation by quadratic formula.

First, we let

a = numerical coefficient of t² = -16
b = numerical coefficient of t = 27
c = constant term = 10

Then, using the quadratic formula

$t = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} = \frac{-27 \pm \sqrt{27^2 - 4(-16)(10)} }{2(-16)} \newline t = \frac{-27 \pm \sqrt{1,369} }{-32} \newline \newline t = \frac{-27 \pm 37 }{32}$

So,

$t = \frac{-27 + 37 }{-32} \text{ or } t = \frac{-27 - 37 }{-32} \newline t = \frac{-10}{32} \text{ or } t = \frac{-64 }{-32} \newline \boxed{ t = -0.3125 \text{ or } t = 2}$

Since t represents the amount of time, t should be positive.

Hence t = 2. Therefore, it takes 2 seconds for the object to hit the ground.

3 0

4 years ago

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Which option correctly represents the graph of f(x) = 1/3x^2+2 and describes

bagirrra123 [75]

The function is even

7 0

3 years ago

What is another way to write 7 8/10 - 4 3/10​

What is another way to write 7 8/10 - 4 3/10