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IRISSAK [1]
3 years ago
11

Answer please please ​

Mathematics
1 answer:
Vlad1618 [11]3 years ago
5 0
The answer is H) the zeros is where it crosses the X axis hope this helps
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Eighteen more than the number n is 125.what is the value of n​
Drupady [299]

Answer: 107

Step-by-step explanation:

6 0
3 years ago
Rewrite the following in the form log (c). 4 log(5)
Murljashka [212]

Answer:

Step-by-step explanation:4log(5) = log(5^4) = log(625).

This problem involves using one of the properties of logs, where a coefficient (in this case the "4") for a logarithm equals the "inside of a logarithm" raised to power of whatever number the coefficient is.

The property in mathematical terms is:    Alog(B) = log(B^A).

So, 4log(5)= log(5^4) = log(625)

5 0
3 years ago
A sphere has a diameter of 12.5 inches. What is the surface area of the sphere? 50π in² 156.25π in² 625π in² 2604.17π in²
Advocard [28]
Hello!

To find the surface area of a sphere you use the equation

A = 4 \pi  r^{2}

A is surface area
r is radius which is half the diameter

Put in the values you know

A = 4 \pi 6.25^{2}

Square the number

A = 4 \pi 39.0625

Multiply the 4 and the 39.0625

A = 156.25 \pi

The answer is B) 156.25 \pi  in^{2}

Hope this helps!
3 0
3 years ago
Problem 4: Solve the initial value problem
pishuonlain [190]

Separate the variables:

y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx

Separate the left side into partial fractions. We want coefficients a and b such that

\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}

\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}

\implies 1 = a(y-2)+b(y+1)

\implies 1 = (a+b)y - 2a+b

\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13

So we have

\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx

Integrating both sides yields

\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx

\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C

\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C

\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C

\dfrac{y-2}{y+1} = e^{3x + C}

\dfrac{y-2}{y+1} = Ce^{3x}

With the initial condition y(0) = 1, we find

\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12

so that the particular solution is

\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}

It's not too hard to solve explicitly for y; notice that

\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}

Then

1 - \dfrac3{y+1} = -\dfrac12 e^{3x}

\dfrac3{y+1} = 1 + \dfrac12 e^{3x}

\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}

y+1 = \dfrac6{2+e^{3x}}

y = \dfrac6{2+e^{3x}} - 1

\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}

7 0
2 years ago
A lifeguard on the shore in a chair that is 6 ft above the ground. She sees a swimmer at an angle of depression of 10°. About ho
melisa1 [442]

Answer:

  34 ft

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you of the relationship between adjacent and opposite sides of a right triangle and the trig function of the associated angle. Here, the relevant relation is ...

  Tan = Opposite/Adjacent

  tan(10°) = (height of lifeguard)/(distance from shore)

Then the distance from shore is ...

  distance from shore = (height of lifeguard)/tan(10°) ≈ (6 ft)/(0.17633)

  distance from shore ≈ 34 ft

3 0
3 years ago
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