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4 years ago

What is .00000023 in scientific notation

Mathematics

1 answer:

Lisa [10]4 years ago

8 0

Answer:

2.3 × 10^-7

Step-by-step explanation:

First, create an equation with the initial decimal being the coefficient multiplied by 10 to the power of 0.

(For example, let’s create the starting scientific notation equation for the decimal 475,000).

475,000 = 475,000 × 10^0

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Now that we have an equation, the second step is to move the decimal point in the coefficient until there is 1 significant digit to the left of the decimal point. For each space the decimal point is moved to the left, increase the exponent by 1.

(Continuing the example above, move the decimal point to the left and increment the exponent.).

475,000 = 475,000.0 × 10^0

475,000 = 47,500.0 × 10^1

475,000 = 4,750.0 × 10^2

475,000 = 475.0 × 10^3

475,000 = 47.5 × 10^4

475,000 = 4.75 × 10^5

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Please help fast Prefer A Expert or Above to Help Me out it’s the bottom one!

erastova [34]

<h3>Answer: 45</h3>

The value is approximate.

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Explanation:

Focus on the upper triangle.

Label the unknown horizontal side as y

Use the pythagorean theorem to solve for y

a^2+b^2 = c^2

10^2 + y^2 = 26^2

100+y^2 = 676

y^2 = 676-100

y^2 = 576

y = sqrt(576)

y = 24

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Now move to the bottom triangle

With respect to the reference angle, the side y = 24 is opposite this. The hypotenuse is unknown.

Use the sine rule to connect the two and solve for x

sin(angle) = opposite/hypotenuse

sin(32) = y/x

sin(32) = 24/x

x*sin(32) = 24

x = 24/sin(32)

x = 45.2899179551967 which is approximate

x = 45 rounding to the nearest whole number

Side x is approximately 45 cm long.

Note: make sure your calculator is in degree mode.

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,