A ball bounces three-fourths the height of its fall. the ball bounces 12ft.,how high does it bounce on the first bounce?

What other information do you need in order to prove the triangles

GuDViN [60]

Answer:

However, we need to know that segment AB is congruent to segment CD, in order to use SAS.

Hope it helps!

Step-by-step explanation:

7 0

3 years ago

A chemist wants to mix a solution of 20% chlorine with a solution of 60% chlorine to make 120 liters of 30% solution of chlorine

Mama L [17]

Answer:

90 and 30 liters

Step-by-step explanation:

1) if solution of 20% is 'x' and of 60% is 'y' liters, then

2) the pure chlorine is: for solution of 20% - 0.2x, for solution of 60% - 0.6y and

3) the mix of the two solutions is 120=x+y.

4) using these items it is possible to make up the system of two equations:

$\left \{ {{x+y=120} \atop {0.2x+0.6y=0.3(x+y)}} \right.$

5) finally, x=90 liters of 20%; y=30 liters of 60% solution.

8 0

2 years ago

A small business assumes that the demand function for one of its new products can be modeled by p = Cekx. When p = $40, x = 1000

Maru [420]

Answer:

C= 82.1116

k=-0.0007192

Step-by-step explanation:

$p=Ce^{kx}\\40=Ce^{k1000}\\30=Ce^{k1400}\\$

Applying logarithmic properties yields in the following linear system:

$ln(40) = ln(C) + 1000k\\ln(30) = ln(C) + 1400k$

Solving for k:

$ln(40) = ln(C) + 1000k\\ln(30) = ln(C) + 1400k\\400 k = ln(30)-ln(40)\\k=-0.0007192$

Solving for C:

$40=Ce^{-0.0007192*1000}\\C= \frac{40}{e^{-0.0007192*1000}}\\C=82.1116$

C= 82.1116

k=-0.0007192

6 0

3 years ago

I need help with the top 3 plz

attashe74 [19]

Answer/Step-by-step explanation:

Recall: SOHCAHTOA

1. Reference angle = 70°

Adjacent side = x

Hypotenuse = 6 cm

Apply CAH. Thus,

Cos 70 = adj/hyp

Cos 70 = x/6

6 × cos 70 = x

2.05 = x

x = 2.05 cm

2. Reference angle = 45°

Adjacent side = x

Hypotenuse = 1.3 m

Applying CAH, we would have the following ratio:

Cos 45 = adj/hyp

Cos 45 = x/1.3

1.3 × cos 45 = x

0.92 = x

x = 0.92 m

3. The who diagram is not shown well. Some parts are missing, however you can still solve the problem just the same way we solved problem 1 and 2.

6 0

3 years ago

Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the

Neko [114]

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees was $x+\frac{1}{4} x=\frac{5}{4} x$ , and for the next three years we have that

Start End

Second year $\frac{5}{4}x$ -------------- $\frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x$

Third year $(\frac{5}{4} )^{2}x$ ------------- $(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x$

Fourth year $(\frac{5}{4})^{3}x$ -------------- $(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.$

So the formula to calculate the number of trees in the fourth year is

$(\frac{5}{4} )^{4} x,$ we know that all of the trees thrived and there were 6250 at the end of 4 year period, then

$6250=(\frac{5}{4} )^{4}x$ ⇒ $x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.$

Therefore the number of trees at the begging of the 4-year period was 2560.

7 0

3 years ago