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3 years ago

HELP ASAP

Mathematics

2 answers:

Schach [20]3 years ago

7 0

Answer:

Step-by-step explanation:

Thanks for my points back

Naily [24]3 years ago

5 0

Answer:

$y+4 = -\frac{1}{3} (x-1)$

Step-by-step explanation:

1) First, find the slope of the line. Use the slope formula $m = \frac{y_2-y_1}{x_2-x_1}$ to find the slope. Substitute the x and y values of the given points into the formula and solve:

$m = \frac{(-5)-(-4)}{(4)-(1)} \\m = \frac{-5+4}{4-1} \\m = \frac{-1}{3}$

So, the slope is $-\frac{1}{3}$ .

2) Use the point-slope formula $y-y_1 = m (x-x_1)$ to write the equation of the line in point-slope form. Substitute values for $m$ , $x_1$ , and $y_1$ into the formula.

Since $m$ represents the slope, substitute $-\frac{1}{3}$ in its place. Since $x_1$ and $y_1$ represent the x and y values of one point on the line, choose any one of the given points (it doesn't matter which one) and substitute the x and y values of it into the formula. (I chose (1,-4), as seen below). This gives the equation of the line in point-slope form.

$y-(-4) = -\frac{1}{3} (x-(1))\\y+4 = -\frac{1}{3} (x-1)$

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Estimate each Quotient

Doss [256]

Okay, through simple division I'll give you the long side of the work. Known as Long Division: [This may turn out unaligned correctly, but hopefully it works.]

Question 1:
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73 / 59628
~~~ 596
~~~ 584 ~ 73 goes into 596 8 times, giving us 584. Subtract and carry.
~~~~ 122
~~~~ []73 ~ 73 goes into 122 1 time, giving us 73. Subtract and carry.
~~~~~ 498
~~~~~ 438 ~ 73 goes into 498 6 times, giving us 438. Subtract and add a 0.
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Question 2:
~~~ [][]6228.35 ~ Answer. Estimate to 6,228.4 or 6,228.
58 / 361245
~~~ 361
~~~ 348 ~ 58 goes into 361 6 times, giving us 348. Subtract and carry.
~~~~ 132
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~~~~~ 116 ~ 58 goes into 164 2 times, giving us 116. Subtract and carry.
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Question 3:
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I hope that helps, have a great rest of your day! ^ ^
{-Ghostgate-}

3 0

3 years ago

Add -4 + (-15).<br><br> -11<br> 19<br> -19<br> 11

chubhunter [2.5K]

Step-by-step explanation:

-4+(-15)

=(-19)

I think it helps you

8 0

3 years ago

In a lab experiment, the decay of a radioactive isotope is being observed. At the beginning of the first day of the experiment t

GaryK [48]

Using an exponential function, it is found that the mass of the radioactive sample at the beginning of the 13th day of the experiment was of 169.1 mg.

<h3>What is an exponential function?</h3>

A decaying exponential function is modeled by:

$A(t) = A(0)(1 - r)^t$

In which:

A(0) is the initial value.

r is the decay rate, as a decimal.

At the beginning of the first day of the experiment the mass of the substance was 500 grams and mass was decreasing by 8% per day, hence A(0) = 500, r = 0.08, and the equation is given by:

$A(t) = A(0)(1 - r)^t$

$A(t) = 500(1 - 0.08)^t$