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3 years ago

12

5. find the missing value (s) so that the line segments are parallel.

1 answer:

Nastasia [14]3 years ago

3 0

The missing value will be (9,5)

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It was John's birthday, 1/2 of his cupcakes were choco lava, the other half was strawberry. There were 10 strawberry cupcakes. R

igomit [66]

If there were 10 strawberry cupcakes then chocolate would have the same number as 10= 1/2. Then 2/4 can be simplified to 1/2 meaning that Rick are 1/2 of 10 and 1/10 is just one out of the 10 strawberry cupcakes.

Rick ate 5 Chocolate Cupcakes

Steward are 1 Strawberry Cupcake

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3 years ago

Read 2 more answers

Help me with this please

Klio2033 [76]

Answer:

No, she's wrong

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

Step-by-step explanation:

So starting from the beginning...

$3x+3\leq -2x+5$

Subtract the 3 from both side and it will look like...

$3x\leq-2x+2$

Then add 2x from both sides and it will look like...

$5x\leq 2$

Finally divide by 5 to both sides and it will look like...

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

4 0

2 years ago

I need help answering

Bas_tet [7]

1/4 c + 3d

c=6 d=7

1/4 (6) + 3(7)

1.5+ 3(7)
1.5+ 21
22.5

5 0

3 years ago

HELP ⚠️⚠️⚠️

Elan Coil [88]

Answer:

the shop meet at March at 65

6 0

3 years ago

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The Laplace Transform of a function f(t), which is defined for all t > 0, is denoted by L{f(t)} and is defined by the imprope

lesya692 [45]

(1) D

$L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt$

Integrate by parts, taking

$u = t \implies \mathrm du=\mathrm dt$

$\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}$

Then

$L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}$

(2) A

$L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}$

7 0

3 years ago