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2 years ago

10

Ms. Jones had $40 to purchase groceries. She bought 4 packs of rice and 3 boxes of granola bars. The packs of rice cost $1.09 ea

ch, and the boxes of granola bars cost $4.29 each. How much money did Ms. Jones have left after these purchases?
answer choices
$8.51

$17.23

$22.77

1 answer:

Nesterboy [21]2 years ago

7 0

Answer:

so to know full price of granola bars and rice packs; you would multiply 1.09 × 4 for the rice which you get 4.36 then for granola bars you do 4.29 × 3 and you get 12.87 then you add 12.87 + 4.36 = 17.23 then you subtract this from the $$ Mr.Jones has which is 40 so its 40-17.23 and you get 22.77

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consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where

boyakko [2]

Answer:

a) $X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)$

b) $Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0 )$

Step-by-step explanation:

From the question we are told that

The Function

$f(x)=1+\frac{1}{x} +\frac{5}{x^2} +\frac{1}{x^3}$

Generally the differentiation of function f(x) is mathematically solved as

$f(x)=1+\frac{1}{x} +\frac{5}{x^2} +\frac{1}{x^3}$

$f(x)=\frac{x^3+x^2+5x+1}{x^2}$

Therefore

$f'(x)=\frac{x^2+10x+3}{x^4}$

Generally critical point is given as

$f'(x)=0$

$\frac{x^2+10x+3}{x^4}=0$

$x=-5 \pm\sqrt{22}$

Generally the maximum and minimum x value for critical point is mathematically solved as

$f'(-5 \pm\sqrt{22})$

Where

Maximum value of x

$f'(-5 +\sqrt{22})$

Minimum value of x

$f'(-5 +\sqrt{22})$

Therefore interval of increase is mathematically given by

$f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})$

$f(x)$

Therefore interval of decrease is mathematically given by

$(-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)$

Generally the second differentiation of function f(x) is mathematically solved as

$f''(x)=\frac{2(x^2+15x+6)}{x^5}$

Generally the point of inflection is mathematically solved as

$f''(x)=0$

$x^2+15x+6=0$

Therefore inflection points is given as

$x=\frac{1}{2} (-15 \pm \sqrt{201}$

$f''(x)>0,\frac{1}{2}(-15-\sqrt{201})$

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$X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)$

$f''(x)$

b)Generally the concave downward interval Y is mathematically given as

$Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0 )$

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2 years ago

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fredd [130]

Answer:

(- 4, - 10 )

Step-by-step explanation:

Given the 2 equations

y = x - 6 → (1)

x = - 4 → (2)

Substitute x = - 4 into (1) for corresponding value of y

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Westkost [7]

Answer:

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Step-by-step explanation:

Given:

$2x^2=5x+6\\\\2x^2-5x-6=0\\\\Where, a=2 , b=-5 , c=-6$

Find:

Value of $x$

Computation:

Using quadratic formula:

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