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2 years ago

Simplify the expression. 1.7 – 2.7 – 1.34 + 4.9 – 1.48

Mathematics

2 answers:

lukranit [14]2 years ago

6 0

Work from left to right
= -1 - 1.34 + 4.9 - 1.48
=-2.34 + 4.9 - 1.48
= 2.56 - 1.48
= 1.08 answer

Lelechka [254]2 years ago

4 0

= -1 - 1.34 + 4.9 - 1.48

=-2.34 + 4.9 - 1.48

= 2.56 - 1.48

= 1.08 answer

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Use a net to find the total surface of the prism.

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Answer:

6682.5cm

Step-by-step explanation:

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Which choice represents the expression below as a single exponential

Lostsunrise [7]

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Step-by-step explanation:

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2 years ago

Find b. Round to the nearest tenth.

Svetradugi [14.3K]

Find Angle B: B = 180 - (82+55) = 43. Then apply the law of sines:

b 8 cm
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2 years ago

Read 2 more answers

Find the value of x (3/5)^x(5/3)^2x=125/27

lesya692 [45]

Given:

The equation is

$\left(\dfrac{3}{5}\right)^x\left(\dfrac{5}{3}\right)^{2x}=\dfrac{125}{27}$

To find:

The value of x.

Solution:

We have,

$\left(\dfrac{3}{5}\right)^x\left(\dfrac{5}{3}\right)^{2x}=\dfrac{125}{27}$

$\left(\dfrac{5}{3}\right)^{-x}\left(\dfrac{5}{3}\right)^{2x}=\dfrac{5\times 5\times 5}{3\times 3\times 3}$ $[\because \left(\dfrac{a}{b}\right)^{-m}=\left(\dfrac{b}{a}\right)^{m}]$

$\left(\dfrac{5}{3}\right)^{-x+2x}=\dfrac{5^3}{3^3}$ $[\because a^ma^n=a^{m+n}]$

$\left(\dfrac{5}{3}\right)^{x}=\left(\dfrac{5}{3}\right)^{3}$

On comparing the exponents, we get

$x=3$

Therefore, the value of x is 3.

8 0

2 years ago

You operate a gaming Web site, www.mudbeast.net, where users must pay a small fee to log on. When you charged $3 the demand was

Doss [256]

Answer:

A) The linear relation between price and demand is:

$d=-550x+2750$

The revenue R is:

$R=-550x^2+2750x$

B) The profit functionP is:

$P=-550x^2+2750x-30$

C) The largest monthly profit is obtained with a log-on fee of $2.5 per month. This corresponds to a profit of $3407.5.

Step-by-step explanation:

We have a site where the number of log-ons depends on our monthly fee. A linear relation is established between the price (log-on fee) and the number of log-ons.

We have two points for this linear relationship:

At price x=3, the demand is d=1100.
At price x=2.5, the demand is d=1375.

We will model the relation:

$d=mx+b$

We can calculate the slope m as:

$m=\dfrac{\Delta d}{\Delta x}=\dfrac{d_2-d_1}{x_2-x_1}=\dfrac{1375-1100}{2.5-3}\\\\\\m=\dfrac{275}{-0.5}=-550$

Then, replacing one point in the linear equation, we can calculate the intercept b:

$d_1=mx_1+b\\\\1100=(-550)\cdot 3+b\\\\1100=-1650+b\\\\b=1100+1650=2750$

Then, the linear relation between demand and price is:

$d=-550x+2750$

The revenue R can be expressed as the multiplication of the price and the demand:

$R=x\cdot d=x(-550x+2750)=-550x^2+2750x$

If we have a fixed cost of $30 per month, the profit P is:

$P=R-FC=-550x^2+2750x-30$

We can maximize the profit by deriving the profit function and making it equal to zero.

$\dfrac{dP}{dx}=0\\\\\\\dfrac{dP}{dx}=-550(2x)+2750(1)=0\\\\\\-1100x+2750=0\\\\x=\dfrac{2750}{1100}=2.5$

This corresponds to a profit of:

$P(2.5)=-550(2.5)^2+2750(2.5)-30\\\\P(2.5)=-550\cdot 6.25+6875-30\\\\P(2.5)=-3437.5+6875-30\\\\P(2.5)=3407.5$

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3 years ago