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3 years ago

Solve the following equation for a. Be sure to take into account whether a letter is

Mathematics

1 answer:

USPshnik [31]3 years ago

6 0

A=6d-g because to be able to get the 6 on the other side you have to multiply 6 on both sides. Then you would have to subtract g to get it on the other side. This causes the gs on the original side to eliminate

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choose a method then find the product 30 times 52

Lunna [17]

First 3x52 = 156 then 156x10 = 1560

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4 years ago

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A new car is purchased for 17900 dollars. The value of the car depreciates at 13% per year. What will the value of the car be, t

Hatshy [7]

17,900 * (1 -0.13)^7 = 6752.86 . . . . dollars

7 0

3 years ago

I need help and i’m trying my best

Cerrena [4.2K]

Answer:

Minimum: 2

Quartile 1: 9

Quartile 2: 11

Quartile 3: 20

Maximum: 24

Step-by-step explanation:

(2,4,{9,10,11,14,20},23,24)

8 0

3 years ago

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Lisa's pet shop has 2 fish tanks.Tank A contains smaller fish who are fed 1 gram of food each per day.Tank B contains 2/3 the nu

Nutka1998 [239]

Answer:

Lisa will not be able to feed both tanks the same amount of food.

Step-by-step explanation:

This question is incomplete, here is the complete question :

"Lisa pet shop has 2 fish tank . Tank A contains smaller fish who were fed 1 gram of food each per day. Tank B contains larger fish who are fed 2 gram each per day. If tank B contains 2/3 the number of fish Tank A contains,will Lisa ever feed both tanks the same amount of food."

Let the number of fish in tank A = f

Lisa fed 1 gram of food each per day = f × 1 = f

So,the number of fish in tank B = $\frac{2}{3}f$

Lisa fed 2 gram of food each per day = $\frac{2}{3}f\times 2$

If both the tanks are fed with same amount of food then

$f=\frac{2f}{3}\times 2$

$f=\frac{4f}{3}$

⇒ 3f = 4f

⇒ 3 = 4

But 3 ≠ 4

Therefore, Lisa will not be able to feed both tanks the same amount of food.

5 0

3 years ago

Differentiate Vx(x + 2) with respect to x.

Travka [436]

Answer:

$y'= \frac{3x+2}{2\sqrt{x} }$

General Formulas and Concepts:

Calculus

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

$y=\sqrt{x} (x+2)$

Step 2: Rewrite

$y=x^{\frac{1}{2} } (x+2)$

Step 3: Differentiate

Product Rule [Basic Power/Chain Rule]: $y'= \frac{1}{2} x^{\frac{1}{2} -1}(x+2)+x^{\frac{1}{2} }(1)$
Simplify: $y'= \frac{1}{2} x^{\frac{-1}{2}}(x+2)+x^{\frac{1}{2}}$
Rewrite: $y'= \frac{x+2}{2\sqrt{x} } + \sqrt{x}$
Add: $y'= \frac{3x+2}{2\sqrt{x} }$

7 0

3 years ago