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2 years ago

Help I need it done now

Mathematics

1 answer:

Paha777 [63]2 years ago

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Can someone help me with 1& 2 will mark brainiest

melamori03 [73]

Sin 45° = opposite side / hypotenuse

1/2 = 8/h
h = 16

2)6/12 = 1/2
1/2 = sin 45°

6√3/12 = √3/2
√3/2 = cos 30°

Sum of angles of triangle = 180°
The 3rd angle = 180 - (45 + 30) = 180 - 75 = 105°

Angles = 30° , 45° , 105°

7 0

2 years ago

7t=x,for t .what is the formula

Andrei [34K]

The formula for t would be x/7.

In order to find this, we simply follow the order of operations to solve for t.

7t = x ----> Divide both sides by 7

t = x/7

6 0

2 years ago

The function in Exercise represents the rate of flow of money in dollars per year. Assume a 10-year period at 8% compounded cont

Kryger [21]

Answer:

a) $P= $56,972.5$

b) $A=$1,26,792.3$

Step-by-step explanation:

Given Data:

Interest rate= $r= 0.08$ per year

No. of years= $t=10$

Rate of continuous money flow is given by the function

$f(t)=2000$

a) to find the present value of money

$P=\int\limits^n_0 {f(t)e^{-rt} } \, dt$

Put f(t)=2000 and n=10 years and r=0.08

$P=\int\limits^n_0 {2000e^{-0.08t} } \, dt$

Now integrate

$P= {2000(\frac{e^{-0.08t}}{-0.08} )$

$P= -\frac{2000}{0.08} (e^{-0.08*10}-e^{-0.08*0})$

$P= -\frac{2000}{0.08} (e^{-0.8}-e^{0})$

$P= -\frac{2000}{0.08} (0.4493-2.7282)$

$P= -\frac{2000}{0.08} (-2.2789)$

$P= -25000(-2.2789)$

$P= $56972.5$

(b) to find the accumulated amount of money at t=10

$A=P(e^{rt} )$

Where P is the present worth already calculated in part a

$A=56972.5(e^{0.08*10} )$

$A=56972.5(e^{0.8} )$

$A=56972.5(2.2255 )$

$A=$1,26,792.3$

5 0

2 years ago

Ginny is studying a population of frogs. She determines that the population is decreasing at an average rate of 3% per year.

Ket [755]

Answer:

The correct answer is C) f(x) = 1,200(0.97)/\x

Step-by-step explanation:

Hope this helps! :)

4 0

2 years ago

List the following numbers in order from least to greatest. <br><br> 2, -3, 9 what is this

scoundrel [369]

-3, 2, 9 hope this helpedd

6 0

2 years ago

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