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2 years ago

7

I need the answers bad

1 answer:

Margarita [4]2 years ago

5 0

Answer:

m<1=72

m<2=72

m<4=72

Step-by-step explanation:

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Write an exponential function in the form y=ab^x that goes through points (0,12) and (3,6144)

skad [1K]

Given:

An exponential function goes through points (0,12) and (3,6144).

To find:

The exponential function.

Solution:

The general exponential function is:

$y=ab^x$ ...(i)

The function goes through point (0,12). Substituting $x=0,y=12$ , we get

$12=ab^0$

$12=a$

The function goes through point (3,6144). Substituting $a=12, x=3,y=6144$ in the general exponential function, we get

$6144=12b^3$

$\dfrac{6144}{12}=b^3$

$512=b^3$

$512^{\frac{1}{3}}=b$

$8=b$

Putting $a=12,b=8$ in (i), we get

$y=12(8)^x$

Therefore, the required exponential function is $y=12(8)^x$ .

3 0

2 years ago

Samantha surveyed 500 of the students in her school about their favorite color.

Alja [10]

Well you start off by doing 27% of 500 which equals 135 then you minus 135 from 500 which leads you to 365 and there is your answer

5 0

2 years ago

Please Solve this, it would be extremely helpful for me.

Sidana [21]

Answer:

Step-by-step explanation:

1) ΔCPD & ΔEPF

∠CPD = ∠EPF { Vertically opposite angles}

∠CDP = ∠PFE {CD║EF, FD is transversal, Alternate interior angles are equal}

ΔCPD ≈ΔEPF {AA criteria for similarity }

$\frac{DC}{EF} =\frac{PC}{EP}\\\\\\\frac{27}{EF}=\frac{15}{7.5}\\\\$

Cross multiply

EF * 15 = 27 * 7.5

$EF =\frac{27*7.5}{15}\\\\$

EF = 27 * 0.5

EF = 13.5 cm

ii) EF // AB, so Triangles ACB & ECF are similar triangles

$\frac{AB}{EF}=\frac{AC}{EC}\\\\\frac{22.5}{13.5}=\frac{AC}{22.5}$

$AC= \frac{22.5*22.5}{13.5}\\\\AC=37.5 cm$

AC = 37.5 cm

6 0

2 years ago

If y= 3x + 6, what is the minimum value of (x^3)(y)?

Rainbow [258]

Answer:

Given the statement: if y =3x+6.

Find the minimum value of $(x^3)(y)$

Let f(x) = $(x^3)(y)$

Substitute the value of y ;

$f(x)=(x^3)(3x+6)$

Distribute the terms;

$f(x)= 3x^4 + 6x^3$

The derivative value of f(x) with respect to x.

$\frac{df}{dx} =\frac{d}{dx}(3x^4+6x^3)$

Using $\frac{d}{dx}(x^n) = nx^{n-1}$

we have;

$\frac{df}{dx} =(12x^3+18x^2)$

Set $\frac{df}{dx} = 0$

then;

$(12x^3+18x^2) =0$

$6x^2(2x + 3) = 0$

By zero product property;

$6x^2=0$ and 2x + 3 = 0

⇒ x=0 and x = $-\frac{3}{2} = -1.5$

then;

at x = 0

f(0) = 0

and

x = -1.5

$f(-1.5) = 3(-1.5)^4 + 6(-1.5)^3 = 15.1875-20.25 = -5.0625$

Hence the minimum value of $(x^3)(y)$ is, -5.0625

3 0

3 years ago

Every month Tanesha pays a fixed fee of $10 to use the parking lot at her workplace. She must also pays $2 each day that she par

Gnom [1K]

10 + 2x • the amount of days in the month

4 0

3 years ago