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3 years ago

Y=5, 12+4y= what is the answer

Mathematics

2 answers:

IRISSAK [1]3 years ago

6 0

Answer: 32

Step-by-step explanation: since y=5, 4y= 20. 12+20=32

Sonbull [250]3 years ago

3 0

Answer:

Step-by-step explanation:

x2 82

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How do you solve 6-(-4)?

mina [271]

Answer:

Step-by-step explanation:

Since you are subtracting a negative number, the negative integer actually becomes positive making the equation 6+4 = 10

5 0

4 years ago

Read 2 more answers

I dont know what to do lol

atroni [7]

Answer:

15 units

Step-by-step explanation:

The distance between two points can be found using the following formula:

$d\sf=\sqrt{(x_2-x_1)+(y_2-y_1)}$

Given points:

(-7, 5) and (2, -7)

<u>where:</u>

x₁ = -7
x₂ = 2
y₁ = 5
y₂ = -7

Substitute these points into the formula and simplify:

$\sf d=\sqrt{(2-(-7))^2+((-7)-5)^2} \ \textsf{[simplify the radicand]}\\\\ d=\sqrt{(2+7)^2+(-12)^2} \ \textsf{[simplify]}\\\\d=\sqrt{(9)^2+(-12)^2} \ \textsf{[evaluate the power]}\\\\d=\sqrt{81+144} \ \textsf{[add]}\\\\d=\sqrt{225}\ \textsf{[simplify]}\\\\d=15$

The distance between the two points is 15 units.

Learn more here:

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8 0

2 years ago

George Kyparisis makes bowling ball in his Miami plant. With recent increases in his cost, he has a newfound interest in efficie

Harlamova29_29 [7]

We compute the productivity percentage change of each category using the formula:

$\text{\% Change }=\frac{\text{ Value now - Value last year}}{\text{Value last year}}\cdot100\%.$

Productivity percentage change for each category

1) Units produced:

$\frac{1,000-1,000}{1,000}\cdot100\%=0\cdot100\%=0\%.$

2) Last (hours)

$\frac{275-300}{300}\cdot100\%=-\frac{25}{300}\cdot100\%=-8.33\%.$

3) Resin (pounded)

$\frac{45-50}{50}\cdot100\%=-\frac{5}{50}\cdot100\%=-10\%.$

4) Capital invested ($)

$\frac{11,000-10,000}{10,000}\cdot100\%=\frac{1,000}{10,000}\cdot100\%=10\%.$

5) Energy (BTU)

$\frac{2,850-3,000}{3,000}\cdot100\%=\frac{-150}{3,000}\cdot100\%=-5\%.$

2) Improvement for labour hours

From the previous computations, we see that the percentage change of the category "Last (hours)" has a negative value of -8.33%, which means that this value decreased by 8.33% with respect to the last year.

Answer

1) Productivity percentage change for each category

0. Units produced: ,0%

1. Last (hours): , -8.33%

2. Resin (pounded):, -10%

3. Capital invested ($): ,10%

4. Energy (BTU): ,-5%

2) Improvement for labour hours

The percentage change of the category "Last (hours)" has a negative value of -8.33%, which means that this value decreased by 8.33% with respect to the last year.

4 0

2 years ago

Referring to the figure, find the value of x

olga2289 [7]

Answer:

x = 183

Step-by-step explanation:

(x -73)° + x° +67° = 360°

x-73+ x +67 = 360

x + x -73 +67 = 360

2x-6 = 360

2x = 360+6

x = 366÷2

x = 183

Check:

(183 - 73) = 110° , x = 183° , 67°

110° + 183° + 67° = 360°

6 0

3 years ago

I have calculus problems that I need help with.

aleksklad [387]

a. Note that $f(x)=x^ne^{-2x}$ is continuous for all $x$ . If $f(x)$ attains a maximum at $x=3$ , then $f'(3) = 0$ . Compute the derivative of $f$ .

$f'(x) = nx^{n-1} e^{-2x} - 2x^n e^{-2x}$

Evaluate this at $x=3$ and solve for $n$ .

$n\cdot3^{n-1} e^{-6} - 2\cdot3^n e^{-6} = 0$

$n\cdot3^{n-1} = 2\cdot3^n$

$\dfrac n2 = \dfrac{3^n}{3^{n-1}}$

$\dfrac n2 = 3 \implies \boxed{n=6}$

To ensure that a maximum is reached for this value of $n$ , we need to check the sign of the second derivative at this critical point.

$f(x) = x^6 e^{-2x} \\\\ \implies f'(x) = 6x^5 e^{-2x} - 2x^6 e^{-2x} \\\\ \implies f''(x) = 30x^4 e^{-2x} - 24x^5 e^{-2x} + 4x^6 e^{-2x} \\\\ \implies f''(3) = -\dfrac{486}{e^6} < 0$

The second derivative at $x=3$ is negative, which indicate the function is concave downward, which in turn means that $f(3)$ is indeed a (local) maximum.

b. When $n=4$ , we have derivatives

$f(x) = x^4 e^{-2x} \\\\ \implies f'(x) = 4x^3 e^{-2x} - 2x^4 e^{-2x} \\\\ \implies f''(x) = 12x^2 e^{-2x} - 16x^3e^{-2x} + 4x^4e^{-2x}$

Inflection points can occur where the second derivative vanishes.

$12x^2 e^{-2x} - 16x^3 e^{-2x} + 4x^4 e^{-2x} = 0$

$12x^2 - 16x^3 + 4x^4 = 0$

$4x^2 (3 - 4x + x^2) = 0$

$4x^2 (x - 3) (x - 1) = 0$

Then we have three possible inflection points when $x=0$ , $x=1$ , or $x=3$ .

To decide which are actually inflection points, check the sign of $f''$ in each of the intervals $(-\infty,0)$ , $(0, 1)$ , $(1, 3)$ , and $(3,\infty)$ . It's enough to check the sign of any test value of $x$ from each interval.