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3 years ago

Plzzzzz help I’ll mark brainlest

Mathematics

2 answers:

irina1246 [14]3 years ago

3 0

When it says find g(-10) it meant when x is -10 what is the solution. By the way the solution is -110.

goblinko [34]3 years ago

3 0

Answer:

-110 is the answer

Hope this helped!!!XD

Also, can I please get brainliest?

Step-by-step explanation:

You might be interested in

Evaluate the expression 4x^4 y^3 when x=1/5 and y=6

wariber [46]

Answer:

5.5296

Step-by-step explanation:

to evaluate the expression 4x^4 y^3 we would substitute the value of x and y into it and evaluate. since x = 1/5 and y = 6

4x^4 y^3

4 × x^4 × 4× y³

4 × (1/5)∧4 × 4 × 6³

4× (0.2)∧4 × 4 × 216

4 × 0.0016 × 864

0.0064 × 864

5.5296

5 0

3 years ago

What's the answer for this question? (The numbers after the letters are indexes btw) 27a9 x 18b5 x 4c2 Over 18a4 x 12b2 x 2c

zysi [14]

Answer:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

Step-by-step explanation:

Given

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Required

Simplify

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Cancel out 18

$\frac{27a^9 * b^5 * 4c^2 }{a^4 * 12b^2 * 2c}$

Divide 4 and 2

$\frac{27a^9 * b^5 * 2c^2 }{a^4 * 12b^2 *c}$

Divide 27 and 12 by 3

$\frac{9a^9 * b^5 * 2c^2 }{a^4 * 4b^2 *c}$

Apply law of indices

$\frac{9a^{9-4} * b^{5-2} * 2c^{2-1} }{4}$

$\frac{9a^5 * b^3 * 2c }{4}$

Divide 2 and 4

$\frac{9a^5 * b^3 * c}{2}$

$\frac{9a^5b^3c}{2}$

Rewrite as:

$\frac{9}{2}a^5b^3c$

Hence:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

4 0

2 years ago

(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti

Rashid [163]

Answer:

The solution

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}$

Step-by-step explanation:

Explanation:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

Step(i):-

Given differential problem

y′+3 y=9 t

Take the Laplace transform of both sides of the differential equation

L( y′+3 y) = L(9 t)

Using Formula Transform of derivatives

L(y¹(t)) = s y⁻(s)-y(0)

By using Laplace transform formula

$L(t) = \frac{1}{S^{2} }$

Step(ii):-

Given

L( y′(t)) + 3 L (y(t)) = 9 L( t)

$s y^{-} (s) - y(0) + 3y^{-}(s) = \frac{9}{s^{2} }$

$s y^{-} (s) - 7 + 3y^{-}(s) = \frac{9}{s^{2} }$

Taking common y⁻(s) and simplification, we get

$( s + 3)y^{-}(s) = \frac{9}{s^{2} }+7$

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

Step(iii):-

By using partial fractions , we get

$\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}$

$\frac{9}{s^{2} (s+3} = \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}$

On simplification we get

9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Put s =0 in equation(i)

9 = B(0+3)

B = 9/3 = 3

Put s = -3 in equation(i)

9 = C(-3)²

C = 1

Given Equation 9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

9 = A s²+3 A s +B(s)+3 B +C(s²)

0 = A + C

put C=1 , becomes A = -1

$\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}$

Step(iv):-

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

$y^{-}(s) =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}$

Applying inverse Laplace transform on both sides

$L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})$

By using inverse Laplace transform

$L^{-1} (\frac{1}{s} ) =1$

$L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}$

$L^{-1} (\frac{1}{s+a} ) =e^{-at}$

Final answer:-

Now the solution , we get

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}$

5 0

3 years ago

Solve for b B=??????

kozerog [31]

Answer:

$\frac{b-7}{b+1}=\frac{b-10}{b+4}$

$\left(b-7\right)\left(b+4\right)=\left(b+1\right)\left(b-10\right)$

$\left(b-7\right)\left(b+4\right)=b^2-3b-28$

$\left(b+1\right)\left(b-10\right)=b^2-9b-10$

$b^2-3b-28=b^2-9b-10$

$b^2-3b-28+28=b^2-9b-10+28$

$b^2-3b=b^2-9b+18$

$b^2-3b-\left(b^2-9b\right)=b^2-9b+18-\left(b^2-9b\right)$

$6b=18$

$\frac{6b}{6}=\frac{18}{6}$

$b=3$

4 0

2 years ago

Use the given values of n and p to find the minimum usual value μ - 2σ and the maximum usual value μ + 2σ. Round your answer to

fenix001 [56]

Given Information:

number of trials = n = 1042

Probability of success = p = 0.80

Required Information:

Maximum usual value = μ + 2σ = ?

Minimum usual value = μ - 2σ = ?

Answer:

Maximum usual value = 859.51

Minimum usual value = 807.78

Step-by-step explanation:

In a binomial distribution, the mean μ is given by

μ = np

μ = 1042*0.80

μ = 833.6

The standard deviation is given by

σ = √np(1 - p)

σ = √1042*0.80(1 - 0.80)

σ = √833.6(0.20)

σ = 12.91

The Maximum and Minimum usual values are

μ + 2σ = 833.6 + 2*12.91

μ + 2σ = 833.6 + 25.82

μ + 2σ = 859.51

μ - 2σ = 833.6 - 25.82

μ - 2σ = 807.78

Therefore, the minimum usual value is 807.78 and maximum usual value is 859.51

8 0

3 years ago