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3 years ago

9

Would appreciate the help with this Math question!

1 answer:

ipn [44]3 years ago

5 0

First option:
c = 2p + 2b

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How does the slope of g(x) compare to the slope of f(x)?

Maru [420]

Answer:

The answer is 'The slope of g(x) is less than the slope of f(x)'

Step-by-step explanation:

Given the graphs of f(x) and g(x). we have to compare the slops of these two.

The graph of f(x) passes through the points (1,0) and (2,2)

∴ $\text{The slope of f(x)=}\frac{y_2-y_1}{x_2-x_1}=\frac{2-0}{2-1}=2$

The graph of g(x) passes through the points (0,2) and (2,3)

∴ $\text{The slope of g(x)=}\frac{y_2-y_1}{x_2-x_1}=\frac{3-2}{2-0}=\frac{1}{2}$

As $\frac{1}{2}$

This shows that the

The slope of g(x) is less than the slope of f(x).

8 0

3 years ago

Read 2 more answers

What are the coordinates of R' for the dilation D(0.5. p)(PQRS)?

11111nata11111 [884]

Answer:

$R' = (-1,-0.5)$

Step-by-step explanation:

See attachment for complete question.

From the attachment:

$P = (-3,-3)$

$R = (1,2)$

Dilation:

$D_{(0.5,P)}$

Required

Determine R'

First, subtract the coordinates of P from R. This means that R is measured from P.

$R = (1-(-3),2-(-3))$

$R = (1+3,2+3)$

$R = (4,5)$

Next, apply dilation factor 0.5

$R' = R * 0.5$

$R' = (4,5) * 0.5$

$R' = (4* 0.5,5* 0.5)$

$R' = (2,2.5)$

Lastly, measure R' from the origin by adding the coordinates of P to R'

$R' = (2+(-3),2.5+(-3))$

$R' = (2-3,2.5-3)$

$R' = (-1,-0.5)$

6 0

3 years ago

1. $1500 earning 7% interest compounded annually for 12 years

ipn [44]

Answer:

$3378.3

Step-by-step explanation:

P= 1500

R= 75

T= 12 years

A= P(1+R/100)^T

A= 1500(1+7/100)^12

A=1500(1.07)^12

A= $3378.3

8 0

3 years ago

Un elevador subió 5 pisos, bajo 9, bajo 12 más, subió 8, bajo otros 4 y se detuvo en el piso 43 ¿De qué piso partió

Alex777 [14]

Answer:

75

Step-by-step explanation:

6 0

3 years ago

Find the exact length of the curve. x = 9 + 9t2, y = 6 + 6t3, 0 ≤ t ≤ 4

andrew11 [14]

ANSWER

$=102 \sqrt{ 17}$

EXPLANATION

The equation of the curve is:

$x = 9 + 9 {t}^{2}$

$y = 6 + 6 {t}^{3}$

We differentiate to obtain:

$\frac{dx}{dt} = 18t$

$\frac{dy}{dt} = 18t$

The length of the arc

$l = \int _0 ^{4} \sqrt{ {( \frac{dx}{dt} )}^{2} + {( \frac{dy}{dt} )}^{2} } dt$

This implies that:

$= \int _0 ^{4} \sqrt{ {( 18t )}^{2} + {( {18 {t}^{2} } )}^{2} } dt$

$= \int _0 ^{4} 18t \sqrt{ 1+ { {18 {t}^{2} } }} dt$

$=102 \sqrt{ 17}$

3 0

3 years ago