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2 years ago

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20) Find the length of a side of a square with a diagonal length of 100 cm.

1 answer:

Andre45 [30]2 years ago

5 0

Answer:

about 7.0710678cm

Step-by-step explanation:

it's a square so the sides are equal.

the diagonal is 100cm

This makes a right triangle with two sides

since the sides are equal, they both are equal to the square root of 50, which is half of 100

the square root of 50 is about 7.0710678cm

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Find the area in the right tail more extreme than z= 3.01 in a standard normal distribution

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P(z > 3.01) = 1 - P(z < 3.01) = 1 - 0.99869 = 0.00131
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2 years ago

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Answer:

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Step-by-step explanation:

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May I ples have a little help pless <br> Simplify using the laws of exponents<br> [2^4/2^ 3]^5

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1 year ago

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muminat

Answer:

(5,-6)

Step-by-step explanation:

ONE WAY:

If $f(x)=x^2-6x+3$ , then $f(x-2)=(x-2)^2-6(x-2)+3$ .

Let's simplify that.

Distribute with $-6(x-2)$ :

$f(x-2)=(x-2)^2-6x+12+3$

Combine the end like terms $12+3$ :

$f(x-2)=(x-2)^2-6x+15$

Use $(x-b)^2=x^2-2bx+b^2$ identity for $(x-2)^2$ :

$f(x-2)=x^2-4x+4-6x+15$

Combine like terms $-4x-6x$ and $4+15$ :

$f(x-2)=x^2-10x+19$

We are given $g(x)=f(x-2)$ .

So we have that $g(x)=x^2-10x+19$ .

The vertex happens at $x=\frac{-b}{2a}$ .

Compare $x^2-10x+19$ to $ax^2+bx+c$ to determine $a,b,\text{ and } c$ .

$a=1$

$b=-10$

$c=19$

Let's plug it in.

$\frac{-b}{2a}$

$\frac{-(-10)}{2(1)}$

$\frac{10}{2}$

$5$

So the $x-$ coordinate is 5.

Let's find the corresponding $y-$ coordinate by evaluating our expression named $g$ at $x=5$ :

$5^2-10(5)+19$

$25-50+19$

$-25+19$

$-6$

So the ordered pair of the vertex is (5,-6).

ANOTHER WAY:

The vertex form of a quadratic is $a(x-h)^2+k$ where the vertex is $(h,k)$ .

Let's put $f$ into this form.

We are given $f(x)=x^2-6x+3$ .

We will need to complete the square.

I like to use the identity $x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2$ .

So If you add something in, you will have to take it out (and vice versa).

$x^2-6x+3$

$x^2-6x+(\frac{6}{2})^2+3-(\frac{6}{2})^2$

$(x+\frac{-6}{2})^2+3-3^2$

$(x+-3)^2+3-9$

$(x-3)^2+-6$

So we have in vertex form $f$ is:

$f(x)=(x-3)^2+-6$ .

The vertex is (3,-6).

So if we are dealing with the function $g(x)=f(x-2)$ .

This means we are going to move the vertex of $f$ right 2 units to figure out the vertex of $g$ which puts us at (3+2,-6)=(5,-6).

The $y-$ coordinate was not effected here because we were only moving horizontally not up/down.

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2 years ago

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3 years ago