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3 years ago

When z=−6, evaluate −2(z+9)−(−5) and −2⋅z−2⋅9+5 to show that −2(z+9)−(−5)=−2⋅z−2⋅9+5

Mathematics

2 answers:

Fantom [35]3 years ago

6 0

Answer:

Yes/true

Step-by-step explanation:

-2(z+9)-(-5)=-2⋅z-2⋅9+5

sostituisco la lettera dandogli un valore

=-2(-6+9)-(-5)=-2⋅(-6)-2⋅9+5

calcolo e tolgo le parentesi

=-2⋅3+5=12-18+5

Elimino i termini uguali

=-2⋅3=12-18

Somma e moltiplicazione tra i numeri

=-6=-6

KengaRu [80]3 years ago

3 0

Ignore this, I posted the wrong answer and I don't know how to delete it.

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MAXImum [283]

Answer:

ohk interesting! indeed

8 0

3 years ago

Use a single digit times a power of 10 to

Nesterboy [21]

Step-by-step explanation:

0.00002468 = 2.468 × 10^(-5)

for single digit =>

2 × 10^(-5)

7 0

2 years ago

Para las celebraciones del barrio de Santiago se junto cierta cantidad de dinero que se distribuirá de la siguiente forma:?

givi [52]

Answer:

4,767

Step-by-step explanation:

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2 years ago

What is the slope of a line that passes through the origin and the point (-2,1)?

Mazyrski [523]

Answer:

The slope would be -1/2.

Step-by-step explanation:

Since, the midpoint of the segment joining the points P (0, -4) and B (8, 0),

= ( 4, -2 ),

Now, the slope of the line passes through the origin and the point ( 4, -2) is,

= -1/2

<h3>i think thats the answer</h3><h2> HOPE IT HELPS YOU THOUGH</h2>

8 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago