All categories

3 years ago

What is the range of the graph (5,3) (2,5) A) 3 < y < 5 B) 3 ≤ y ≤ 5 C) 2 < x < 5 D) 2 ≤ x ≤ 5

Mathematics

1 answer:

Tresset [83]3 years ago

7 0

Answer:

Step-by-step explanation:

domain is the set of x-values

range is the set of y-values (corresponding to the x-values)

Now, since we want the range, we can immediately eliminate Choices C & D because they dealing with the DOMAIN ONLY (since x values).

Now, we have to choose from option A or option B.

The only difference is the greater than or less than sign with "equal to" signs as well. Whenever we are given 2 coordinate points, we can say that ALWAYS, we can graph a line segment using these 2 points. And line segments are all SOLID {meaning that the endpoints are INCLUDED).

What we mean by included is that the inequality sign should have "equal to" with it. Also, the two values of y-values are 3 and 5, thus range would be between 3 and 5 (inclusive). We can write:

Range = 3 ≤ y ≤ 5

Correct choice is B

You might be interested in

Strain-displacement relationship) Consider a unit cube of a solid occupying the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 After loa

Anastasy [175]

Answer:

please see answers are as in the explanation.

Step-by-step explanation:

As from the data of complete question,

$0\leq x\leq 1\\0\leq y\leq 1\\0\leq z\leq 1\\u= \alpha x\\v=\beta y\\w=0$

The question also has 3 parts given as

Part a: Sketch the deformed shape for α=0.03, β=-0.01 .

Solution

As w is 0 so the deflection is only in the x and y plane and thus can be sketched in xy plane.

the new points are calculated as follows

Point A(x=0,y=0)

Point A'(x+αx,y+βy)

Point A'(0+(0.03)(0),0+(-0.01)(0))

Point A'(0,0)

Point B(x=1,y=0)

Point B'(x+αx,y+βy)

Point B'(1+(0.03)(1),0+(-0.01)(0))

Point B'(1.03,0)

Point C(x=1,y=1)

Point C'(x+αx,y+βy)

Point C'(1+(0.03)(1),1+(-0.01)(1))

Point C'(1.03,0.99)

Point D(x=0,y=1)

Point D'(x+αx,y+βy)

Point D'(0+(0.03)(0),1+(-0.01)(1))

Point D'(0,0.99)

So the new points are A'(0,0), B'(1.03,0), C'(1.03,0.99) and D'(0,0.99)

The plot is attached with the solution.

Part b: Calculate the six strain components.

Solution

Normal Strain Components

$\epsilon_{xx}=\frac{\partial u}{\partial x}=\frac{\partial (\alpha x)}{\partial x}=\alpha =0.03\\\epsilon_{yy}=\frac{\partial v}{\partial y}=\frac{\partial ( \beta y)}{\partial y}=\beta =-0.01\\\epsilon_{zz}=\frac{\partial w}{\partial z}=\frac{\partial (0)}{\partial z}=0\\$

Shear Strain Components

$\gamma_{xy}=\gamma_{yx}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=0\\\gamma_{xz}=\gamma_{zx}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}=0\\\gamma_{yz}=\gamma_{zy}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=0$

Part c: Find the volume change

$\Delta V=(1.03 \times 0.99 \times 1)-(1 \times 1 \times 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\$

Also the change in volume is 0.0197

For the unit cube, the change in terms of strains is given as

$\Delta V={V_0}[(1+\epsilon_{xx})]\times[(1+\epsilon_{yy})]\times [(1+\epsilon_{zz})]-[1 \times 1 \times 1]\\\Delta V={V_0}[1+\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}+\epsilon_{xx}\epsilon_{zz}+\epsilon_{yy}\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}\epsilon_{zz}-1]\\\Delta V={V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the strain values are small second and higher order values are ignored so

$\Delta V\approx {V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\ \Delta V\approx [\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the initial volume of cube is unitary so this result can be proved.

5 0

2 years ago

Which zero pair could be added to the function so that the function can be written in vertex form?

Oduvanchick [21]

The function in this problem should be: f(x) =x² + 12x + 6

y = x² + 12x + 6
y - 6 = x² + 12x

x² ⇒ x * x
12x ⇒ 2*6*x
missing number is 6² = 36

y - 6 + 36 = x² + 12x + 36

(x+6)(x+6) ⇒ x(x+6)+6(x+6) ⇒ x² + 6x + 6x + 36 = x² + 12x + 36

y + 30 = x² + 12x + 36
y = (x+6)² - 30

Choice is D. 36,-36

5 0

3 years ago

What is the complement and supplement of 74°?

serious [3.7K]

I think it's 90 degrees

5 0

3 years ago

Read 2 more answers

The owners of an eyeglass store claim that they fill customers' orders, on average, in 60 minutes with a standard deviation of 1

melomori [17]

By applying z-score formula we got that if a reporter determines a mean order time of 63 minutes then value o Z is 1.19

<h3>What is standard deviation?</h3>

Standard deviation is a number used to tell how measurements for a group are spread out from the average (mean), or expected value

Here given that

The owners of an eyeglass store claim that they fill customers' orders, on average, in 60 minutes with a standard deviation of 15. 9 minutes. Based on a random sample of 40 orders, a reporter determines a mean order time of 63 minutes

mean=μ=60

standard deviation=σ=15.9

number of sample=n=40

$\bar{x}=63 \\$

Now we can calculate z by z-score formula as

$z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \\\\&=\frac{63-60}{\frac{15 \cdot 9}{40}} \\ \\=\frac{3}{2.5140} \\ \\=1.19\end{aligned}$

By applying z-score formula we got that if a reporter determines a mean order time of 63 minutes then value o Z is 1.19

To learn more about standard deviation visit :brainly.com/question/12402189

5 0

2 years ago

CAN SOMEONE PLEASE HELP ME SOLVE THIS I WILL GIVE U BRAINIEST PLEASE AND THANK YOU.!

Arisa [49]

Answer:

Step-by-step explanation:

SIMPLE 30 X 30

8 0

2 years ago