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3 years ago

Please help???? Geometry

Mathematics

1 answer:

Free_Kalibri [48]3 years ago

7 0

Find the slope=(y2-y1)/(x2-x1)

m=(0-3)/(0--1/2)

m=-3/(1/2)

m=-3(2/1)

m=-6 so

y=-6x+b, since we have (0,0)

0=-6(0)+b

0=0+b

0=b

y=-6x

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Simplify 20÷5+8^2×2-9

nevsk [136]

20/5+8^2*2-9

20/5+64*2-9

4+64*2-9

4+ 128-9

132-9

123

I hope this helped!
~cupcake

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3 years ago

Read 2 more answers

How do you solve | -23.12 + .2361 | + 31.876 ?

anzhelika [568]

Answer:

54.7599

Step-by-step explanation:

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2 years ago

A cylindrical water tank has a radius of 6 feet and a height of 20 feet. A larger tank with the same height has a volume that is

likoan [24]

Answer:

Radius of large cylinder $= 10.39 feet\\$

Step-by-step explanation:

Volume of the larger tank is three times the volume of smaller tank

Volume of smaller tank

$= \pi r^{2} h\\= (3.14) * 6^{2} * 20\\= 2260.8 ft^{3} \\$

Volume of larger tank

$= 3 * 2260.8 ft^3\\= 6782.4 ft^3\\$

Or it can also be written as

$\pi r^{2} h = 6782.4 ft^3\\(3.14) * ( r^2) 20 = 6782.4 ft^3\\r^2 = 108\\r = 10.39$

6 0

3 years ago

A swimming pool whose volume is 10 comma 000 gal contains water that is 0.03% chlorine. Starting at tequals0, city water conta

Katarina [22]

Answer:

C(60) = 2.7*10⁻⁴

t = 1870.72 s

Step-by-step explanation:

Let x(t) be the amount of chlorine in the pool at time t. Then the concentration of chlorine is

C(t) = 3*10⁻⁴*x(t).

The input rate is 6*(0.001/100) = 6*10⁻⁵.

The output rate is 6*C(t) = 6*(3*10⁻⁴*x(t)) = 18*10⁻⁴*x(t)

The initial condition is x(0) = C(0)*10⁴/3 = (0.03/100)*10⁴/3 = 1.

The problem is to find C(60) in percents and to find t such that 3*10⁻⁴*x(t) = 0.002/100.

Remember, 1 h = 60 minutes. The initial value problem is

dx/dt= 6*10⁻⁵ - 18*10⁻⁴x = - 6* 10⁻⁴*(3x - 10⁻¹) x(0) = 1.

The equation is separable. It can be rewritten as dx/(3x - 10⁻¹) = -6*10⁻⁴dt.

The integration of both sides gives us

Ln |3x - 0.1| / 3 = -6*10⁻⁴*t + C or |3x - 0.1| = e∧(3C)*e∧(-18*10⁻⁴t).

Therefore, 3x - 0.1 = C₁*e∧(-18*10⁻⁴t).

Plug in the initial condition t = 0, x = 1 to obtain C₁ = 2.9.

Thus the solution to the IVP is

x(t) = (1/3)(2.9*e∧(-18*10⁻⁴t)+0.1)

then

C(t) = 3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴t)+0.1)

If t = 60

We have

C(60) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴*60)+0.1) = 2.7*10⁻⁴

Now, we obtain t such that 3*10⁻⁴*x(t) = 2*10⁻⁵

3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 2*10⁻⁵

t = 1870.72 s

7 0

3 years ago

A grocery store manager claims that 75% of shoppers purchase bananas as least once a month. Technology was used to simulate choo

Keith_Richards [23]

Answer:

P-value = 0.0333

At 5% level of significance;

0.0333 < 0.05

Therefore, we reject null hypothesis H₀ at 5% level of significance,

We conclude that proportion of shoppers who bought bananas at least once in the past month is overstated

Step-by-step explanation:

Given the data in the question;

To test whether population proportion p is overstated;

Null hypothesis H₀ : p = (75%) = 0.75

Alternative hypothesis H₁ : = < (75%) < 0.75

now, sample proportion p" = 64 / 100 = 0.64

from the dot plot below, we will determine the p-value for test { P(p" < 0.64)}

so, the number of times p"<0.64 in 150 simulations is 5

Hence; P(p" < 0.64 ) = 5 / 150 = 0.0333

P-value = 0.0333

At 5% level of significance;

0.0333 < 0.05

Therefore, we reject null hypothesis H₀ at 5% level of significance,

We conclude that proportion of shoppers who bought bananas at least once in the past month is overstated

7 0

2 years ago