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galben [10]
3 years ago
5

Need help hurry please

Mathematics
1 answer:
shusha [124]3 years ago
6 0
The slope is the number next to the variable and the y intercept is the number after the slope.

#1. Slope= -3
Y-intercept=-3

#2. Slope=5/2
Y-intercept=-4

#3. Slope=1/3
Y-intercept=3

#4. Slope=-2
Y-intercept=2

#5. Slope=-2/5
Y-intercept=-2

#6. Slope=-2/3
Y-intercept=-2
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What is the distributive property of 10n+6
Dahasolnce [82]
You can rewrite 10n + 6 as 2(5n + 3). 
Find a common factor in both 10 and 6, which is 2 and factor the 2 out. 2 times <em>5</em> gives 10 and 2 times <em>3</em> gives 6.
6 0
3 years ago
Need help with this Q&amp;A
enot [183]

Answer:

the second option

y = (-1/2)x + 11

Step-by-step explanation:

x = 2

y = (-1/2)×2 + 11 = -1 + 11 = 10

the same as the table.

x = 4

y = (-1/2)×4 + 11 = -2 + 11 = 9

the same as the table.

x = 6

y = -3 + 11 = 8

the same as the table.

x = 8

y = -4 + 11 = 7

the same as the table.

8 0
3 years ago
Quadrilateral ABCD is an Isosceles Trapezoid. Which angle is congruent to Angle C?
weqwewe [10]
Answer: Angle D

The base angles C and D are congruent because this is an isosceles trapezoid. In addition, angle A = angle B because this is another pair of base angles (imagine rotating the figure 180 degrees to make segment AB the base)
3 0
3 years ago
Read 2 more answers
Create a variable expression that represents the word phrase: twelve times a number
ASHA 777 [7]
12n or 12
\times n
7 0
3 years ago
According to one cosmological theory, there were equal amounts of the two uranium isotopes 235U and 238U at the creation of the
FromTheMoon [43]

Answer:

6 billion years.

Step-by-step explanation:

According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let P(t) be the amount of ^{235}U and Q(t) be the amount of ^{238}U after t years.

Then, we obtain two differential equations

                               \frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q

where k_1 and k_2 are proportionality constants and the minus signs denotes decay.

Rearranging terms in the equations gives

                             \frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt

Now, the variables are separated, P and Q appear only on the left, and t appears only on the right, so that we can integrate both sides.

                         \int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt

which yields

                      \ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2,

where c_1 and c_2 are constants of integration.

By taking exponents, we obtain

                     e^{\ln |P|} = e^{-k_1t + c_1}  \quad e^{\ln |Q|} = e^{-k_12t + c_2}

Hence,

                            P  = C_1e^{-k_1t} \quad Q  = C_2e^{-k_2t},

where C_1 := \pm e^{c_1} and C_2 := \pm e^{c_2}.

Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition

                                 P(0) = Q(0) = C

Substituting 0 for P in the general solution gives

                         C = P(0) = C_1 e^0 \implies C= C_1

Similarly, we obtain C = C_2 and

                                P  = Ce^{-k_1t} \quad Q  = Ce^{-k_2t}

The relation between the decay constant k and the half-life is given by

                                            \tau = \frac{\ln 2}{k}

We can use this fact to determine the numeric values of the decay constants k_1 and k_2. Thus,

                     4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}

and

                     7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}

Therefore,

                              P  = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q  = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}

We have that

                                          \frac{P(t)}{Q(t)} = 137.7

Hence,

                                   \frac{Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} }{Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}} = 137.7

Solving for t yields t \approx 6 \times 10^9, which means that the age of the  universe is about 6 billion years.

5 0
3 years ago
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