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frez [133]
3 years ago
12

Danielle earns a 7.75% commission on everything she sells at the electronics store where she works. She also earns a base salary

of $700 per week. How much did she earn last week if she sold $4,000 in electronics merchandise? Round your intermediate calculations and answer to the nearest cent.
Mathematics
1 answer:
In-s [12.5K]3 years ago
3 0

y = .0775x + 700

y = .0775(4,000) + 700

y = 310 + 700

y = 1,010

danielle earned $1,010 last week if she sold $4,000 in electronics merchandise.

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What is the approximate volume of one refraction cup? What’s its formula ?
Marizza181 [45]

Answer:

Volume = 0.1696 L

The formula is Area of base of refraction cup × Depth of the refraction cup

Step-by-step explanation:

The dimensions of a refraction cup are;

Diameter = 120 mm

Depth = 30 mm

Therefore, volume of the refraction cup = Area of base × Depth

Volume of the refraction cup = (π×120²/4)/2 × 30 = 169646.003 mm³

1 L = 1000000 mm³

Therefore, 169646.003 mm³ = 169646.003/1000000 m³ = 0.1696 L

The formula is Area of base of refraction cup × Depth of the refraction cup.

6 0
2 years ago
Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)
ELEN [110]

Question:

Find the point (,) on the curve y = \sqrt x that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})

Step-by-step explanation:

y = \sqrt x can be represented as: (x,y)

Substitute \sqrt x for y

(x,y) = (x,\sqrt x)

So, next:

Calculate the distance between (x,\sqrt x) and (3,0)

Distance is calculated as:

d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}

So:

d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}

d = \sqrt{(x-3)^2 + (\sqrt x)^2}

Evaluate all exponents

d = \sqrt{x^2 - 6x +9 + x}

Rewrite as:

d = \sqrt{x^2 + x- 6x +9 }

d = \sqrt{x^2 - 5x +9 }

Differentiate using chain rule:

Let

u = x^2 - 5x +9

\frac{du}{dx} = 2x - 5

So:

d = \sqrt u

d = u^\frac{1}{2}

\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}

Chain Rule:

d' = \frac{du}{dx} * \frac{dd}{du}

d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}

d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}

d' = \frac{2x - 5}{2\sqrt u}

Substitute: u = x^2 - 5x +9

d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}

Next, is to minimize (by equating d' to 0)

\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0

Cross Multiply

2x - 5 = 0

Solve for x

2x  =5

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Substitute x = \frac{5}{2} in y = \sqrt x

y = \sqrt{\frac{5}{2}}

Split

y = \frac{\sqrt 5}{\sqrt 2}

Rationalize

y = \frac{\sqrt 5}{\sqrt 2} *  \frac{\sqrt 2}{\sqrt 2}

y = \frac{\sqrt {10}}{\sqrt 4}

y = \frac{\sqrt {10}}{2}

Hence:

(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})

3 0
3 years ago
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serg [7]

Answer:

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Step-by-step explanation:

From the information,

s+3=c

(1/2)c=g

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juin [17]

Answer:

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so

(q^4)^z = q^(4*z)

now

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so

z = 3

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Leto [7]

Answer:

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Step-by-step explanation:

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