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2 years ago

(0, 2), (1,7) create a equation that passes through each pair of points

Mathematics

1 answer:

uranmaximum [27]2 years ago

8 0

Answer:

y = 5x + 2

Step-by-step explanation:

Given the equation 5x + 2

Plug in 0

y = 5(0) + 2

y = 2

(0,2)

Given the equation 5x + 2

Plug in 1

y = 5(1) + 2

y = 7

(1,7)

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Carys calculates the total amount EEE, in dollars, that she earns for working hhh hours using the equation E=10hE=10hE, equals,

timurjin [86]

Answer:

$10 per hour
0.1 hours

Step-by-step explanation:

To find Carys' earnings in one hour, fill in h=1 in the formula:

E = 10·1 = 10

Carys earns 10 dollars per hour.

To find the number of hours required to earn 1 dollar, fill in E=1 in the formula:

1 = 10h

0.1 = h . . . . . divide by 10

It takes 0.1 hours for Carys to earn 1 dollar.

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3 years ago

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How am i Invisible???????????????/

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Answer:

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Step-by-step explanation:

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3 years ago

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You work for a company that produces custom picture frames. A new customer needs to frame a piece of rectangular artwork with di

miskamm [114]

Answer:

None of the choices are correct.

Step-by-step explanation:

Let x be the width of the frame,

The framed art cannot me more than 320 square inches

$(11 + 2x) \times (15+2x) \leq 320$

$(165 +30x +22x +4x^2 ) \leq 320$

$4x^2+52x + 165 \leq 320$

$4x^2 +52x + 165-320 \leq 0$

$4x^2 +52x - 155\leq 0$

By using the quadratic formula

$x = \frac{-b\pm \sqrt{(b^2-4ac)}}{2a}$

$x = \frac{-52\pm \sqrt{(52^2-4(4)(-155))}}{2(4)}$

$x = \frac{-52\pm \sqrt{(2704+2480)}}{2(4)}$

$x = \frac{-52\pm \sqrt{(5184)}}{2(4)}$

$x = \frac{-52\pm 72}{2(4)}$

$x = \frac{20}{8}$

$x= \frac{5}{2}$

$x = 2.5$

Frame must not be more than 2.5 inches wide.

4 0

3 years ago

What is the missing value in this table of equivalent ratios?

Hitman42 [59]

Answer:

Step-by-step explanation:

follow the pattern in its timetables,hope this helps u!

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3 years ago

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

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3 years ago