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3 years ago

[PLEASE HELP! WILL GIVE BRAINLIEST IF CORRECT!]

Mathematics

2 answers:

natita [175]3 years ago

5 0

A just trust me :-> if your wrong I’m sorry :

tankabanditka [31]3 years ago

3 0

<h2>Answer:</h2>

Option: C is the correct answer.

C: x-axis: minutes in increments of 1; y-axis: temperature in increments of 1

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

As we know that x-axis is used to denote the independent variable and y-axis is used to represent the dependent variable.

In the given question the temperature is recorded over a period of time.

This means that the x-axis will represent the time in minutes and y-axis will represent the temperature.

Also, we are given a table as:

Minutes Temperature (°C)

0 -2

1 1

2 3

3 4

4 5

5 -4

6 2

7 -3

Hence, we see that the minutes are increasing by every 1 unit and the temperature also has a very small change and are integers, so we will give a increment of 1 in the y-axis as well.

By doing so the graph would be clear and easily understood.

Hence, the answer is:

C: x-axis: minutes in increments of 1; y-axis: temperature in increments of 1

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Rearrange the equation so w is the independent variable.<br> u-5=-4(w-1)

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$u-5=-4(w-1) \\u=5-4(w-1)\\u=5-4w+4\\u=-4w+9$

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3 years ago

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$