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3 years ago

A flower has ten pedals.During a storm,five of the pedals fall off.Write a decimal to show what part of the pedals is left

Mathematics

1 answer:

olga nikolaevna [1]3 years ago

7 0

The answer is .5 meaning half (1/2) of the petals fell off.

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Write 337,060 in expanded form using exponents

Agata [3.3K]

1. expand using place value.

337,060 = 300,000 + 30,000 + 7,000 + 000 + 60 + 0

Next, we will use exponents:

300,000 = 3 * 10^5

30,000 = 3 * 10^4

7,000 = 7 * 10^3

000 = 0 * 10^2

60 = 6 * 10

0 = 0 * 10^0

then after combing these exponents, we can write the number as:

337,060 = 3 * 10^5 + 3 * 10^4 + 7 * 10^3 + 0 * 10^2 + 6 * 10 + 0 * 10^0

Finally, removing the meaningless zeroes, we would end up with:

337,060 = 3 * 10^5 + 3 * 10^4 + 7 * 10^3 + 6 * 10

3 0

3 years ago

One card is drawn form a standard deck of 52 cards. What is the probability of <br> P(jack or red)

bazaltina [42]

2/52 which simplifies down to 1/26

7 0

4 years ago

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Given: SY and TZ, transversal UX ∠YVW and _______ are alternate interior

Bond [772]

< yvw and < twv are alternate interior angles

opposite sides of the transversal and inside are alternate interior angles

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3 years ago

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Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago

X intercept f(x)=(x+4)^2

natali 33 [55]

Answer:

Step-by-step explanation:

Given the function f(x)=(x+4)², <u>the x intercept occurs at f(x) = 0,</u> substituting f(x) = 0 into the given equation to get the value of x we have;

$0 = (x+4)^{2} \\$

Taking the square root of both sides

$\sqrt{(x+4})^{2} = \sqrt{0}\\ x+4 = 0\\$

Subtracting 4 from both sides of the equation

$x+4-4 = 0-4\\x = -4$

Therefore the x intercept of the equation is -4

4 0

3 years ago