What is the point on the terminal side of -210

Consider the relationship between monthly electricity bill for a household during the summer and average high ambient temperatur

grigory [225]

Answer: True

Step-by-step explanation:

a variable (often denoted by y ) whose value depends on that of another. The temperature is depending on the electricity bill

7 0

3 years ago

Factor to find the zeros of the function defined by the quadratic expression. −7x2 − 91x − 280

Mariulka [41]

There are no zeroes for an expression, only for equations.
Assuming equation to be
−7x^2 − 91x − 280=0
-7(x^2+13x+40)=0
-7(x+8)(x+5)=0
by the zero product properties,
x+8=0 => x=-8
or
x+5=0 => x=-5

3 0

3 years ago

It's even better if you have a telescope or binoculars. Both allow you to see planets, the moon, and stars even closer. Telescop

babunello [35]

Answer:

A

Step-by-step explanation:

hope i helped

5 0

3 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

3 years ago

Two large and 1 small pumps can fill a swimming pool in 4 hours. One large and 3 small pumps can also fill the same swimming poo

kolezko [41]

Let x and y be the unit rates at which one large pump and one small pump works, respectively.

Two large/one small operate at a unit rate of

(1 pool)/(4 hours) = 0.25 pool/hour

so that

2x + y = 0.25

One large/three small operate at the same rate,

(1 pool)/(4 hours) = 0.25 pool/hour

x + 3y = 0.25

Solve for x and y. We have

y = 0.25 - 2x ==> x + 3 (0.25 - 2x) = 0.25

==> x + 0.75 - 6x = 0.25

==> 5x = 0.5

==> x = 0.1

==> y = 0.25 - 2 (0.1) = 0.25 - 0.2 = 0.05

In other words, one large pump alone can fill a 1/10 of a pool in one hour, while one small pump alone can fill 1/20 of a pool in one hour.

Now, if you have four each of the large and small pumps, they will work at a rate of

4x + 4y = 4 (0.1) + 4 (0.05) = 0.6

meaning they can fill 3/5 of a pool in one hour. If it takes time t to fill one pool, we have

(3/5 pool/hour) (t hours) = 1 pool

==> t = (1 pool) / (3/5 pool/hour) = 5/3 hours

So it would take 5/3 hours, or 100 minutes, for this arrangement of pumps to fill one pool.

6 0

3 years ago

What is the point on the terminal side of -210​

What is the point on the terminal side of -210