Could the table x,y pairs have been generated by a linear function x 1,2,3,4 y 3,5,7,9

1+1x5

Lesechka [4]

Answer:

10 eheyyahwhwhahwhwhwhwhwhhe

4 0

3 years ago

Read 2 more answers

he graph shows f(x) = [1/2]^x and its translation, g(x). Which describes the translation of f(x) to g(x)? translation of four un

Annette [7]

Answer by YourHope:

Hi! :)

The graph shows f(x) = [1/2]^x and its translation, g(x). Which describes the translation of f(x) to g(x)?

Translation of four units up! The f(x) function cannot possibly be f(x)=1/2, though, because that would be a horizontal line through y = 1/2 and that function is clearly not a horizontal line. So whatever f(x) is really, add 4 to the tail end of it to show its translation!

:)

6 0

4 years ago

Solve the discriminant

Marizza181 [45]

Answer:

a

Step-by-step explanation:

given a quadratic equation in standard form

ax² + bx + c = 0 ( a ≠ 0 )

then the discriminant

Δ = b² - 4ac

• if b² - 4ac > 0 then 2 real solutions

• if b² - 4ac = 0 then 2 real and equal solutions

• if b² - 4ac < 0 then no real solutions

given

$\frac{3}{4}$ x² - 3x = - 4 ( add 4 to both sides )

$\frac{3}{4}$ x² - 3x + 4 = 0 ← in standard form

with a = $\frac{3}{4}$ b = - 3 , c = 4

then

b² - 4ac = (- 3)² - ( 4 × $\frac{3}{4}$ × 4) = 9 - 12 = - 3

since b² - 4ac < 0 then equation has no real solutions

5 0

2 years ago

The floor of a conference hall can be covered completly with tiles.Its length is 36ft. longer than its width.The area of the flo

Masteriza [31]

Answer:

Step-by-step explanation:

Given

Area of a floor = 2040ft²

If Its length is 36ft. longer than its width, then L = W+36

Area = LW

L is the length

W is the width

A < W(W+36)

2040 < W²+36W

0<W²+36W-2040

W²+36W-2040 >0

W = -36±√36²-4(-2040)/2

W = -36±√1296+8160)/2

W = -36±√9456/2

W = -36±97.25/2

W = -36+97.25/2

W = 61.24/2

W > 30.62ft

Since L = W+36

L > 30.62+36

L > 66.62ft

The width of the floor is 30.62ft

The length is 66.62ft

The mathematical sentence would represent the given situation is the width is greater than 30.62ft while the length must be greater than 66.62ft

The possible dimension of the floor is 31ft by 67 ft

The possible areas is 31*67 = 2077ft²

It won't be realistic to get an area of 144sqft because the initial area is greater than 144. Hence the feasible area will be greater than 2040ft²

5 0

3 years ago

How does one integrate the following:

grin007 [14]

Consider substituting $u=4+xy$ , so that $\mathrm du=x\,\mathrm dy$ . Then

$\displaystyle\int_{x=a}^{x=b}\int_{y=a}^{y=b}\frac x{(4+xy)^2}\,\mathrm dy\,\mathrm dx=\int_{x=a}^{x=b}\int_{u=4+ax}^{u=4+bx}\frac1{u^2}\,\mathrm du\,\mathrm dx$
$=\displaystyle\int_{x=a}^{x=b}\left(-\frac1u\right)\bigg|_{u=4+ax}^{u=4+bx}\,\mathrm dx$
$=\displaystyle\int_{x=a}^{x=b}\left(\frac1{4+ax}-\frac1{4+bx}\right)\,\mathrm dx$

Then by similar substitutions, you can easily find that you end up with

$\dfrac1a\ln|4+ax|-\dfrac1b\ln|4+bx|\bigg|_{x=a}^{x=b}$
$=\dfrac1a\ln|4+ab|-\dfrac1b\ln|4+b^2|-\dfrac1a\ln|4+a^2|+\dfrac1b\ln|4+ab|$
$=\dfrac1a\ln\left|\dfrac{4+ab}{4+a^2}\right|+\dfrac1b\ln\left|\dfrac{4+ab}{4+b^2}\right|$

Of course, this all assumes that the integrand is continuous over the domain of integration, which would require that $a,b$ are chosen such that $xy\neq-4$ for any $(x,y)\in[a,b]^2$ . If in particular $ab>-4$ , then we can write

$=\dfrac1a\ln\dfrac{4+ab}{4+a^2}+\dfrac1b\ln\dfrac{4+ab}{4+b^2}$

and you can combine the logarithms if you like as

$=\ln\sqrt[a]{\dfrac{4+ab}{4+a^2}}+\ln\sqrt[b]{\dfrac{4+ab}{4+b^2}}$
$=\ln\left(\sqrt[a]{\dfrac{4+ab}{4+a^2}}\sqrt[b]{\dfrac{4+ab}{4+b^2}}\right)$

7 0

3 years ago