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umka2103 [35]
2 years ago
10

Solve a/-1.2 = -0.4. a=?

Mathematics
1 answer:
adell [148]2 years ago
6 0

Answer:

a = 0.48

Step-by-step explanation:

Multiply all terms by the same value to eliminate fraction denominators

then simplify

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Answer: 23

Step-by-step explanation: Because you do 34-11

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A caterer for a wedding reception wants to use a recipe for eggplant salad that recommends using 1.8 kg of eggplant for 10
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<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%20lnx%20%7D%20" id="TexFormula1" title=" \sqrt[4]{ lnx } " alt=" \sqrt[4]
Kamila [148]

Use the power rule for differentiation:

\dfrac{\text{d}}{\text{d}x} (f(x))^k = k(f(x))^{k-1}f'(x)

You can use this formula if you remember that a root is just a rational exponential:

\sqrt[4]\ln(x) = (\ln(x))^{\frac{1}{4}}

So, remembering that the derivative of the logarithm is 1/x, you have

\dfrac{\text{d}}{\text{d}x} (\ln(x))^{\frac{1}{4}} = \frac{1}{4}(\ln(x))^{\frac{1}{4}-1}\dfrac{1}{x}

Which you can rewrite as

\dfrac{1}{4}(\ln(x))^{\frac{1}{4}-1}\dfrac{1}{x} =\dfrac{1}{4}(\ln(x))^{\frac{-3}{4}}\dfrac{1}{x} =\dfrac{1}{4}\dfrac{1}{\sqrt[4]{\ln(x))^3}}\dfrac{1}{x} = \dfrac{1}{4x\sqrt[4]{\ln(x))^3}}

3 0
3 years ago
Back Next
emmasim [6.3K]

Answer:

C

Step-by-step explanation:

✔️First, solve for r:

r/2 ≤ 3

Multiply both sides by 2

r/2 × 2 ≤ 3 × 2

r ≤ 6

This implies that possible value of r is equal to 6 or less than 6.

Graphing this on a number line, the line with a shaded circle, indicating that 6 is included, starts at 6 and points to the left.

This indicates that value of r ranges from 6 and below.

The graph is C.

7 0
3 years ago
Solve this quadratic equation.<br><br> 6x2 - 2x = 7
rjkz [21]

Answer:

x₁ = (1 + √43)/6

x₂ = (1 - √43)/6

Step-by-step explanation:

6x² - 2x = 7

6x² - 2x - 7 = 0

Formula

ax² + bx + c = 0

a = 6 ; b = - 2 ; c = - 7

x₁/₂ = [-b ± √(b²- 4ac)]/2a

x₁/₂ = (2 ± √4 + 4×6×-7)/12

x₁/₂ = (2 ± √172)/12

x₁/₂ = (2 ± 2√43)/12

x₁/₂ = 2(1 ± √43)/12

x₁/₂ = (1 ± √43)/6

x₁ = (1 + √43)/6

x₂ = (1 - √43)/6

5 0
3 years ago
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