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3 years ago

Help me please picture is attached

Mathematics

2 answers:

wlad13 [49]3 years ago

7 0

Girl i can barley see that

Hoochie [10]3 years ago

6 0

I can’t see it send me a nether picture!!

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Pls help! What is the range in interval notation?

Anon25 [30]

Do you love Selena Quintanilla?

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3 years ago

Mr.Church had 3 muffins he sold 2 but then baked 6 more and sold 4 more how many muffin did he not sell.

9966 [12]

He failed to sell three muffins

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3 years ago

Read 2 more answers

Can u solve the answer

pickupchik [31]

Answer:

She needs 50 ft² of wood to cover the table

Step-by-step explanation:

You need to know the formula for area of a trapezoid, which is

A = (B1 + B2)h/s (add the bases together, multiply by the height, then divide

by 2)

We are given B1 = 12, B2 = 8, and h = 5, so plug them in and simplify

A = (12 + 8)(5)/2

A = (20)(5)/2

A = 100/2

A = 50 ft²

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3 years ago

Mike and Kim invest $18,000 in equipment to print yearbooks for school. Each yearbook costs $5 to print and sells for $15. How m

dsp73

Sorry... you probably don't need the answer now but It is d...

I don't know who reported the last person but they are correct

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4 years ago

The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%

liq [111]

<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by $\frac{100+1.8}{100}$ = $\frac{101.8}{100}$ .

So, the population in the year t can be given by $P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}$

Population in the year 2000 = $3,381,000\textrm{x}(\frac{101.8}{100})^{6}$ = $3,762,979.38$

Population in year 2000 = 3,762,979

Let us assume population doubles by year $y$ .

$2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}$

$log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})$

$y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537$

$y$ ≈ $2033$

∴ By 2033, the population doubles.

4 0

4 years ago