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Zielflug [23.3K]

2 years ago

9

The test results for 4 students are: 96, 83, 78, and 83. If one more student's test score of 87 is added, what would increase? H

elp please!!!

1 answer:

Alex2 years ago

5 0

Well the average would change from 85 to 85.4

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Find the measure of 0. Round answer to the nearest tenth. Pleas help

DENIUS [597]

Answer:

$\theta=36.43^{\circ}$

Step-by-step explanation:

Given that,

Base = 42

Perpendicular height = 31

We need to find the value of $\theta$ . It can be calculated using trigonometry as follows :

$\tan\theta=\dfrac{P}{B}\\\\\theta=\tan^{-1}(\dfrac{P}{B})$

Put all the values,

$\theta=\tan^{-1}(\dfrac{31}{42})\\\\\theta=36.43^{\circ}$

So, the value of $\theta=36.43^{\circ}$ .

6 0

2 years ago

I’m confused please help

Maru [420]

<h2>Hello!</h2>

The answer is:

$Sin(A)=0.93$

<h2>Why?</h2>

To solve the problem, remember to set your calculator in "degree mode"

So, we are given that the angle A is equal to 68°

So, we have that:

$Sin(A)=Sin(68\°)=0.93$

Hence, we have that:

$Sin(A)=0.93$

Have a nice day!

4 0

2 years ago

On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes

belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

$r = \frac{\Delta T_{X}}{\Delta T_{Y}}$

$r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}$

$r = 11\,\frac{^{\circ}X}{^{\circ}Y}$

The difference between current temperature in Y linear scale with respect to freezing point is:

$\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)$

$\Delta T_{Y} = 115\,^{\circ}Y$

The change in X linear scale is:

$\Delta T_{X} = r\cdot \Delta T_{Y}$

$\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)$

$\Delta T_{X} = 1265\,^{\circ}X$

Lastly, the current temperature on the X scale is:

$T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X$

$T_{X} = 1150\,^{\circ}X$

The current temperature on the X scale is 1150 °X.

5 0

2 years ago

.<br><br> y = x2 + 3<br> y = x + 5

ddd [48]

I assume you mean x squared in the first equation

Because both are equal to y, they are equal to each other so x + 5 = x^2 +3
If we then move everything over to one side, we get x^2 - x - 2 = 0

Then factorise it to (x-2)(x+1) = 0
And solve both parts separately

x + 1 = 0
x = -1

x-2 = 0
x = 2

Sub both values into the simplest equation in this case y=x+5
to get y = 4 and y = 7

7 0

1 year ago

Identify the nonlinear equation. A) y = x B) y = 3x - 7 C) y = x2 D) y = 3 E) y = 1 3

Softa [21]

ANSWER

C)

$y = {x}^{2}$

EXPLANATION

A linear function is a function whose graph is a straight line.

The graph of y=x is a straight line.

The graph of y=3x -7 is also a straight line.

The graph of y=x² is a parabola.

This graph is not a straight line .

It is a non-linear function.

y=3 and y=13 are also straight lines.

The correct choice is C.

7 0

3 years ago