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3 years ago

Please help asap! Leaf plot

Mathematics

1 answer:

Over [174]3 years ago

3 0

The answer is B. A Stem and Leaf Plot is a special table where each data value is split into a "stem" which is the first digit of each number and a "leaf" the last number of each digit.

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On what interval (s) are f(x) = x3 2x²+x+6 decreasing?

sammy [17]

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6 0

3 years ago

Read 2 more answers

The minimum and maximum temperature on a cold day in Lollypop Town can be modeled by the equation below: 2|x − 6| + 14 = 38 What

enyata [817]

$2|x-6|+14=38$
$\rightarrow2|x-6|=24$
$\rightarrow|x-6|=12$

There are two cases:
1. $x-6=12\rightarrow x=18$
2. $x-6=-12\rightarrow x=-6$

So the minimum temperature is -6, and the maximum is 18.

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3 years ago

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Enter an algebraic expression to model the given context. Give your answer in simplest form. The original price p of an item les

ra1l [238]

Answer:

Step-by-step explanation:

Let

Original price = p

Less Discount = 20%

Total discount paid for the item = 20% of p

= 20/100 * p

= 0.2 * p

= 0.2p

Actual price after discount = 100% - 20%

= 80%

Total price paid for the item = 80% of p

= 80/100 * p

= 0.8 * p

= 0.8p

6 0

4 years ago

Polynomial Equation question (Screenshot included)

Illusion [34]

$y = {x}^{3} - 4 {x}^{2} - 45x \\ y = x( {x}^{2} - 4x - 45) \\ y = x(x + 5)(x - 9) \\ x = 0, - 5,9$
So, to find our x-intercepts, y must be zero
$0 = x \\ \\ 0 = (x + 5) \\ - 5 = x \\ \\ 0 = (x - 9) \\ 9 = x$
ANSWER: (0,0) (-5.0) (9,0) are your x-intercepts

6 0

4 years ago

What is the function written in vertex form?

lys-0071 [83]

Answer:

The answer in the procedure

Step-by-step explanation:

The question does not present the graph, however it can be answered to help the student solve similar problems.

we know that

The equation of a vertical parabola into vertex form is equal to

$f(x)=a(x-h)^{2}+k$

where

a is a coefficient

(h,k) is the vertex

If the coefficient a is positive then the parabola open up and the vertex is a minimum

If the coefficient a is negative then the parabola open down and the vertex is a maximum

case A) we have

$f(x)=3(x+4)^{2}-6$

The vertex is the point (-4,-6)

a=3

therefore

The parabola open up, the vertex is a minimum

case B) we have

$f(x)=3(x+4)^{2}-38$

The vertex is the point (-4,-38)

a=3

therefore

The parabola open up, the vertex is a minimum

case C) we have

$f(x)=3(x-4)^{2}-6$

The vertex is the point (4,-6)

a=3

therefore

The parabola open up, the vertex is a minimum

case D) we have

$f(x)=3(x-4)^{2}-38$

The vertex is the point (4,-38)

a=3

therefore

The parabola open up, the vertex is a minimum

4 0

3 years ago