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3 years ago

What relationships exist between the figures on the coordinate grid below?

Mathematics

1 answer:

lbvjy [14]3 years ago

5 0

Answer: the relations would be where the line meets the corner of a square where both x and y have a number

Step-by-step explanation:

if you showed a picture of the graph it would help, but where the line of solutions meets a corner

You might be interested in

The slope of a line parallel to -x + y = 12 equals what?

Viktor [21]

1. parallel lines have same slope

7 0

3 years ago

A pilot can fly 2100 miles with the wind in the same time she can fly 1890 miles against the wind. If the speed of the wind is 2

Anna007 [38]

Answer:

475 mph

Step-by-step explanation:

Distance = rate × time

time = distance / rate

If x is the speed of the plane in still air:

2100 / (x + 25) = 1890 / (x − 25)

Cross multiply:

2100 (x − 25) = 1890 (x + 25)

Distribute and solve:

2100x − 52500 = 1890x + 47250

210x = 99750

x = 475

5 0

3 years ago

How many terms are in the arithmetic sequence shown below? 15, 7, -1, -9...,-225

NemiM [27]

We know that

We can write an Arithmetic Sequence as a rule:

an = a1 + d(n−1)

where

a1 = the first term
d =the "common difference" between terms

in this problem
a1=15 a2=7 a3=-1 a4=-9 ..... an=-225
d=a2-a1
d=7-15-----> d=-8

an = a1 + d(n−1)
for
an=-225
d=-8
a1=15
find n
-225=15+(-8)*(n-1)--> (n-1)=[-225-15]/-8----> n-1=30---> n=30+1---> n=31

the answer is
31

3 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

4/12×1/8 in simplest form

Sliva [168]

Answer:

$\frac{1}{24}$

Step-by-step explanation:

$\frac{4}{12} *\frac{1}{8} =\frac{1}{24}$

6 0

3 years ago