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3 years ago

9

For Sue’s party, her family rented a bouncy house for $150 plus $2 per child, c.

1 answer:

Vladimir79 [104]3 years ago

7 0

Jairo
Hsiii. Kiwibanqikq

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Let f(x)=x2−8x+19<br><br> What is the minimum value of the function?

nataly862011 [7]

By completing the square: x²-8x+19 = (x-4)²-16+19 = (x-4)²+3 so minimum point = (4,3)

By calculus: f'(x) = 2x-8. (minimum point occurs when f'(x)=0 therefore)
2x-8 = 0 means x=4 and the corresponding y value is 3.

The minimum value of this function (y) is 3 and this occurs when x = 4.

7 0

3 years ago

Read 2 more answers

Somebody help sad face I don’t like test

Nana76 [90]

Answer:

1 3d

Step-by-step explanation:

The Way That you cross it out is a half -1 and a cross is 1 3d

8 0

2 years ago

Amanda want to pay the same amount each month on her car loan. The number of payments varies inversely with the monthly payment

Oxana [17]

Answer:

Her monthly payment would be $280.

Step-by-step explanation:

Same amount of $ per month

175 times 48 equals 8,400

(175 = monthly payment, 48 = amount of months)

8,400 divided by 30 equals 280

(8,400 = total amount of money, 30 = amount of months)

This makes the months = money ratio:

30 months = $280.

7 0

3 years ago

What is the equation of the line that passes through (6, 1) and is perpendicular to the line y 3x+2? CI) At what angle do the li

Andrews [41]

Answer with explanation:

The equation of any line with slope 'm' and passing through any point $(x_1,y_1)$ is given by

$y=mx+(y_1-mx_1)$

As we know that the general equation of a line with slope 'm' is $y=mx+c$

Comparing with the given equation $y=3x+2$ we can conclude slope of the given line is $m_1=3$

Now we know that the product of slopes of perpendicular lines is -1

Mathematically we can write for perpendicular lines

$m_1\times m_2=-1$

Thus the slope of the required line is obtained from the above relation since it is given that they are perpendicular

$3\times m_{2}=-1\\\\\therefore m_{2}=\frac{-1}{3}$

Hence using the given and the obtained values the equation of the required line is

$y=-\frac{1}{3}x+(1-\frac{-1}{3}\times 6)}\\\\y=\frac{-x}{3}+3$

Part b)

The angle of intersection between 2 lines with slopes $m_{1},m_2$ is given by

$\theta =tan^{-1}(\frac{m_2-m_1}{1+m_1m_2})$

Comparing the equations of given lines

$y=6x+3\\\\y=2x+3$

with the standard equation we get

$m_{1}=6\\\\m_{2}=2$

Thus the angle of intersection becomes

$\theta =tan^{-1}(\frac{2-6}{1+2\times6})=17.10$

8 0

3 years ago

Can someone answer the top question? Thanks!

oee [108]

Answer:

Least to greatest:

x-1

x

1-x

-1/x

5 0

3 years ago