Please please help with these two !!! i’ll give brainliest pls

Find the difference. express the answer in scientific notation. (5.29 times 10 superscript 11 baseline) minus (3.86 times 10 sup

Georgia [21]

The difference between (5.29 times 10 superscript 11 baseline) minus (3.86 times 10 superscript 11 baseline) is 1. 43 × 10^11

<h3>How to determine the notation</h3>

Given the expression

(5. 29 × 10^11) - (3. 86 × 10 ^11)

First, find the common factor

10^11 ( 5. 29 - 3. 86)

Then substract the values within the bracket

10^11 (1. 43)

Multiply with the factor, we have

⇒1. 43 × 10^11

Thus, the difference between (5.29 times 10 superscript 11 baseline) minus (3.86 times 10 superscript 11 baseline) is 1. 43 × 10^11

Learn more about index notation here:

brainly.com/question/10339517

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7 0

1 year ago

Who kno the answer ? Please help

Aleks04 [339]

It's the last one. .........

4 0

3 years ago

Read 2 more answers

4. Which of the following is equal to (9 x 1) + (5 x 1/10) + (1 x 1/100)?

Stolb23 [73]

Answer:

c

Step-by-step explanation:

ccccccccccccccccccccccccc

5 0

3 years ago

According to a census performed in a certain country in a particular year, about 209,280,000 of it's residents were adults, and

gayaneshka [121]

54160000/209280000

= 0.2587
=25.88%

7 0

3 years ago

How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

1 year ago