Which is the shorts side of the triangle?

Write an equation you could use for this word problem. Do not need to solve.

blondinia [14]

Answer:

23 + x = 40

Step-by-step explanation:

Hi,

23 is the total amount your friend has. 23 + the total amount you have is equal to 40. Since we don't know how much you have, we can write x. Therefore, 23 + x = 40 would be correct.

Hope this helps :)

7 0

2 years ago

Read 2 more answers

Match The Term

GalinKa [24]

Answer:

jyyj

Step-by-step explanation:

From the figure,

Radius matches with (F) that is a segment between the centre of a circle and a point on the circle.

Arc matches with (D) that is Circle A and the portion of the circle between points B and C is darkened.

Chord matches with (C) Circle A and a line segment connecting points B and C which are on the circle.

Secant matches with (E) that is Circle A and a line segment connecting points B and C which are on the circle.

Tangent matches with (A) that is a line intersects a circle in exactly one point.

Circumference matches (B) that is the distance around a circle.

4 0

3 years ago

Read 2 more answers

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

Use the image below to answer the following question. What relationship do the ratios of sin x° and cos y° share?

Drupady [299]

Answer:

B

Step-by-step explanation:

$y \times {4 \div 14 \frac{10}{4 |25 \\ | } }^{2}$

8 0

2 years ago

Use the expression 14X + 28Y + 21.. what is the greatest common factor

12345 [234]

Answer:

1

Step-by-step explanation:

None of these numbers are alike terms so the only number left is 1.

4 0

2 years ago