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3 years ago

Which is the same as moving the decimal point 3 places to the right in a decimal number

Mathematics

1 answer:

BlackZzzverrR [31]3 years ago

8 0

Answer:

Moving the decimal 3 places to the right in a decimal number is the same as multiplying the number by 1000.

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A line passes through the points (-1, -6) and (-4, 3). What is the equation of this line in slope-intercept form?

iVinArrow [24]

Answer:

Step-by-step explanation:

Slope = (3+6)/(-4+1)

= 9/-3

= -3

Y=mx+b

-6= -3×-1 +b

-6 = 3 +b

b = -9

6 0

3 years ago

Are the ratios 3:6 and 1:2 equivalent?

aalyn [17]

Answer:

yes these are equivalent bcz

3/6=1/2

Step-by-step explanation:

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3 years ago

Read 2 more answers

there are 158 4th grade students and 139 fifth grade students going on a field trip .there needs to be at least 1 adult chaperon

strojnjashka [21]

Answer:

(D) is the best explanation

Step-by-step explanation:

The total number of students is 158 + 139 = 297. This number can be rewritten as ...

297 = 37 × 8 + 1

That is, if there were 37 chaperones, 1 student would remain without a chaperone. That means the number of chaperones must be increased by 1 to 38. This is best explained by choice D.

3 0

2 years ago

What does a residual value of 0.8 mean in reference to the line of best fit?

notka56 [123]

Answer:

i think its -The given point is 0.8 units above the line of best fit

Step-by-step explanation:

only because if it were -0.8 it would be below the line

6 0

2 years ago

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We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the c

Kisachek [45]

We can solve this problem using separation of variables.

Then apply the initial conditions

EXPLANATION

We were given the first order differential equation

$\frac{dT}{dt}=k(T-a)$

We now separate the time and the temperature variables as follows,

$\frac{dT}{T-a}=kdt$

Integrating both sides of the differential equation, we obtain;

$ln(T-a)=kt +c$

This natural logarithmic equation can be rewritten as;

$T-a=e^{kt +c}$

Applying the laws of exponents, we obtain,

$T-a=e^{kt}\times e^{c}$

$T-a=e^{c}e^{kt}$

We were given the initial conditions,

$T(0)=4$

Let us apply this condition to obtain;

$4-20=e^{c}e^{k(0)}$

$-16=e^{c}$

Now our equation, becomes

$T-a=-16e^{kt}$

or

$T=a-16e^{kt}$

When we substitute a=20,
we obtain,

$T=20-16e^{kt}$

b) We were also given that,

$T(5)=8$

Let us apply this condition again to find k.

$8=20-16e^{5k}$

This implied

$-12=-16e^{5k}$

$\frac{-12}{-16}=e^{5k}$

$\frac{3}{4}=e^{5k}$

We take logarithm to base e of both sides,

$ln(\frac{3}{4})=5k$

This implies that,

$\frac{ln(\frac{3}{4})}{5}=k$

$k=-0.2877$

After 15 minutes, the temperature will be,

$T=20-16e^{-0.2877\times 15}$

$T=20-0.21376$

$T=19.786$

After 15 minutes, the temperature is approximately 20°C

6 0

3 years ago