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3 years ago

Pls help need asap in 10 mins

Mathematics

1 answer:

CaHeK987 [17]3 years ago

4 0

The last one. In arithmetic sequence, the difference between the terms is constant, in this case, the difference is 3

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Solve: 2x-3 = 5(x + 6)

Slav-nsk [51]

Answer:

2x - 3 = 5(x + 6)

2x - 3 = 5x + 30

-33 = 3x

x = -11

7 0

2 years ago

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Explain how you can rewrite the equation n+8=24 so that it involves subtraction rather than addiction

Vinil7 [7]

Answer:

n = 24 - 8.

Step-by-step explanation:

n + 8 = 24

Subtract 8 from each side:

n + 8 - 8 = 24 - 8

n = 24 - 8.

8 0

3 years ago

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Find the length of the curve y = 3/5x^5/3 - 3/4x^1/3 + 6 for 1 < = x < = 8. The length of the curve is . (Type an exact an

Mashutka [201]

Answer:

$\sqrt\frac{387}{20}$

Step-by-step explanation:

$Arc Length =\int\limits^a_b {\sqrt{1+(\frac{dy}{dx})^2 } } \, dx$

$y=\dfrac{3}{5}x^{\frac{5}{3}}- \dfrac{3}{4}x^{\frac{1}{3}}+6$

$\frac{dy}{dx} =x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}}$

$1+(\frac{dy}{dx})^2 }=1+(x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}})^2\\=1+(x^{\frac{4}{3}}-\dfrac{1}{2}+ \dfrac{1}{16}x^{-\frac{4}{3}})$

$=\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}}$

For the Interval $1\leq x\leq 8$

Length of the Curve $=\int\limits^8_1 {\sqrt{\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}} } } \, dx\\$

Using T1-Calculator

$=\sqrt\frac{387}{20}$

3 0

3 years ago

What is the probability that an offspring will have a homozygous dominant genotype?

Anton [14]

Answer:

The probability is 1/2% or 50%

Step-by-step explanation:

3 0

3 years ago

After releasing of radioactive material into.The atnosphere from.A nuclear.Powe plant in a country in 1988, the hay in that coun

Radda [10]

Answer:

the question is incomplete, so I looked for a similar one:

<em>After the release of radioactive material into the atmosphere from a nuclear power plant, the hay was contaminated by iodine 131 ( half-life, 8 days). If it is all right to feed the hay to cows when 10% of the iodine 131 remains, how long did the farmers need to wait to use this hay? </em>

iodine's half life (we are given x, we need to find b):

0.5A₀ = A₀eᵇˣ

x = 8 days

we eliminate A₀ from both sides

0.5 = eᵇ⁸

ln 0.5 = ln eᵇ⁸

-0.69315 = b8

b = -0.69315 / 8 = -0.08664

since the farmers need to wait until only 10% of the iodine remains (we already calculated b, now we need to find x):

0.1A₀ = A₀eᵇˣ

0.1 = eᵇˣ

ln 0.1 = bx

where b = -0.08664

x = ln 0.1 / -0.08664 = -2.302585 / -0.08664 = 26.58 days

4 0

3 years ago