All categories

3 years ago

Helppp pleaseeee!!!!

Mathematics

1 answer:

Savatey [412]3 years ago

4 0

Answer:

Initial value = 4

Growth factor = 1.6

Step-by-step explanation:

Function representing the exponential growth is given by,

f(x) = Initial value(1 + growth rate)ˣ

Here x = time or duration for growth

(1 + growth rate) = Growth factor

Given function is,

f(x) = 4(1.6)ˣ

By comparing both the functions,

Initial value = 4

Growth factor = 1.6

You might be interested in

Fred says that 1/2 and 7/8 are equivalent fractions. Draw area models for 1/2 and 7/8 to show if Fred's statement is correct. Na

jok3333 [9.3K]

Answer:

For your area models you would draw two rectangles and divide one into 2 equal sections, and the other one with 8 equal sections. You then shade 1 box out of the first triangle with 2 equal sides. Then you would shade 7 boxes on the rectangle with 8 equal sides. These diagrams would show you that 7/8 and 1/2 are not equivalent.

To find out what fractions are equivalent to 1/2 you would take 1/2 and times the numerator and denominator by the same number.

1/2*2/2= 2/4

2/4=1/2

1/2*9/9= 9/18

9/18=1/2

2/4 and 9/18 both equal 1/2

Hope this helps ;)

4 0

3 years ago

These are so hard for me

aleksandr82 [10.1K]

Answer:

Step-by-step explanation:

The range is the values of y from the vertex upwards

The vertex = (- 2, - 4) → y = - 4

The range is from - 4 upwards in the direction of the parabola, that is

range is [ - 4, ∞ )

6 0

3 years ago

Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!

IgorC [24]

Hey there, hope I can help!

$-4x^2+9y^2+32x+36y-64=0$

$\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \ 9y^2+32x+36y-4x^2=64$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \ -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4$
$-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \ -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1$

$Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1$

For me I used
$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1$
$As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}$

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
$\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1$

Hope this helps!

4 0

3 years ago

You are picking up a friend and going to a concert. your friend lives 3 miles west of your house and the concert is 21 miles eas

artcher [175]

24 miles away from your house

3 0

3 years ago

nikitadnepr [17]

The answer is all real numbers

5 0

3 years ago