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2 years ago

Solve Highschool Pre cal

Mathematics

1 answer:

Bezzdna [24]2 years ago

6 0

Option C:

$\ln 64 y^{2}$

Solution:

Given expression: 2 ln 8 + 2 ln y

Applying log rule $a \log b=\log b^{a}$ in the above expression.

⇒ $2 \ln 8+2 \ln y=\ln 8^{2}+\ln y^{2}$

Applying log rule $\log a+\log b=\log (a b)$ in the above expression.

⇒ $\ln 8^{2}+\ln y^{2}=\ln \left(8^{2} y^{2}\right)$

We know that $8^{2}=64$

⇒ $\ln \left(8^{2} y^{2}\right)=\ln 64 y^{2}$

Hence, the expression in single natural logarithm is $\ln 64 y^{2}$ .

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A. You can get cash-back rewards

Step-by-step explanation:

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2 years ago

Which ordered pair is a solution to the system of equations below? y = -2x – 1 y = 3x – 26

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2 years ago

Matthew is going to drive from Salt Lake City, Utah, to Las Vegas, Nevada. His car travels a consistent

Eddi Din [679]

Answer: 14 gallons of gasoline.

Step-by-step explanation:

given data:

25(x + 3) = 424

where x is the number of gallon of gasoline needed to purchase.

Solution.

25(x + 3) = 424

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2 years ago

Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)

Sergeeva-Olga [200]

Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

$x^{2} +y^{2}+4x-1=0$

Step-by-step explanation:

Explanation:-

Step 1:-

The equation of the circle having center and radius is

$(x-h)^2+(y-k)^2=r^2$

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

$(\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2} )$

$(-2,0)$

therefore center (h,k) = (-2,0)

Step 2:-

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

$\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}$

$r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}$

$r=\sqrt{5}$

Step 3:-

center (h,k) = (-2,0) and

radius $r=\sqrt{5}$

The standard form of circle equation

$(x-h)^2+(y-k)^2=r^2$

$(x-(-2))^2+(y-0)^2=\sqrt{5} ^2$

on simplification is

$x^{2} +y^{2}+4 x-1=0$

5 0

2 years ago

The graph of f is given in the figure to the right. Let A(x)equals=Integral from 0 to x f left parenthesis t right parenthesis

Tpy6a [65]

Answer:

$A(4)=-4\pi$

$A(8)=-4\pi +8$

$A(12)=-4\pi +16$

$A(14)=-4\pi +15$

Step-by-step explanation:

we are given

$A(x)=\int\limits^x_0 f{x} \, dx$

Calculation of A(4):

we can plug x=4

$A(4)=\int\limits^4_0 f{x} \, dx$

Since, this curve is below x-axis

so, the value of integral must be negative

and it is quarter of circle

so, we can find area of quarter circle

radius =4

$A(4)=-\frac{1}{4}\times \pi \times (4)^2$

$A(4)=-4\pi$

Calculation of A(8):

we can plug x=8

$A(8)=\int\limits^8_0 f{x} \, dx$

we can break into two parts

$A(8)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx$

now, we can find area and then combine them

$A(8)=-4\pi +\frac{1}{2}\times 4\times 4$

$A(8)=-4\pi +8$

Calculation of A(12):

we can plug x=12

$A(12)=\int\limits^12_0 f{x} \, dx$

we can break into two parts

$A(12)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx+\int\limits^12_8 f{x} \, dx$

now, we can find area and then combine them

$A(12)=-4\pi +\frac{1}{2}\times 8\times 4$

$A(12)=-4\pi +16$

Calculation of A(14):

we can plug x=14

$A(14)=\int\limits^14_0 f{x} \, dx$

we can break into two parts

$A(14)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx+\int\limits^12_8 f{x} \, dx+\int\limits^14_12 f{x} \, dx$

now, we can find area and then combine them

$A(14)=-4\pi +\frac{1}{2}\times 8\times 4-\frac{1}{2}\times 1\times 2$

$A(14)=-4\pi +16-1$

$A(14)=-4\pi +15$

8 0

2 years ago