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3 years ago

Solve for X (24x – 19) (16х + 13)

Mathematics

1 answer:

worty [1.4K]3 years ago

6 0

Step-by-step explanation:

24x - 19 = 16x + 13

24x = 16x + 32

8x = 32

x = 4

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There are four nickels and seven dimes in your pocket. One of the nickels and one of the dimes are Canadian. The others are US c

Lana71 [14]

Answer:

$P=0.727$

Step-by-step explanation:

The total coins in your pocket are 11, the probability of a nickel coming out from a random selection is $\frac{4}{11}$ , and the probability of taking out a dime is $\frac{7}{11}$ . In the case of the Canadian currency, you have two Canadian coins of eleven, the probability is $\frac{2}{11}$ . The probability of select a dime or a Canadian coin is the sum of the probability of each event less the probability of selecting the coin that is a dime and a Canadian coin:

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4 years ago

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4 years ago

Find the constant difference for a hyperbola with foci (-3.5, 0) and (3.5, 0) and a point on the hyperbola (3. 5, 24).

Solnce55 [7]

Answer:

2a=1

Step-by-step explanation:

The constant difference for a hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is $2a.$

1. Since hyperbola has foci (-3.5, 0) and (3.5, 0), then $c=3.5.$ Note that $c=\sqrt{a^2+b^2},$ then

$\sqrt{a^2+b^2}=3.5\Rightarrow a^2+b^2=3.5^2.$

2. Since point (3.5,24) lies on the hyperbola, then

$\dfrac{3.5^2}{a^2}-\dfrac{24^2}{b^2}=1.$

3. Solve the system of two equations:

$\left\{\begin{array}{l}a^2+b^2=3.5^2\\\dfrac{3.5^2}{a^2}-\dfrac{24^2}{b^2}=1\end{array}\right.$

From the 1st equation,

$b^2=3.5^2-a^2,$

then

$\dfrac{3.5^2}{a^2}-\dfrac{24^2}{3.5^2-a^2}=1,\\ \\\dfrac{3.5^2(3.5^2-a^2)-24^2a^2}{a^2(3.5^2-a^2)}=1,\\ \\3.5^4-3.5^2a^2-24^2a^2=3.5^2a^2-a^4,\\ \\a^4-a^2(2\cdot 3.5^2+24^2)+3.5^4=0,\\ \\a^4-600.5a^2+150.0625=0,\\ \\D=(-600.5)^2-4\cdot 150.0625=360000,\\ \\a^2_{1,2}=\dfrac{600.5\pm 600}{2}=\dfrac{1}{4},600\dfrac{1}{4}.$