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postnew [5]
2 years ago
11

Which shows the correct placement of the decimal point in the product?

Mathematics
1 answer:
kolbaska11 [484]2 years ago
5 0

Answer:

Step-by-step explanation:

177.14 Glad to help  :D

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Question 10(Multiple Choice Worth 1 points) (06.01 LC) Choose the polynomial that is written in standard form.
Svetach [21]

Answer:

  −3x^8 + 9x^2 + 10x

Step-by-step explanation:

A polynomial is in standard form when the exponents of the variable decrease left to right. The only given expression in that form is ...

  −3x^8 + 9x^2 + 10x

7 0
2 years ago
Given that
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-6 This is your answer it help you

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3 years ago
Which expression should be used to estimate the answer to this story problem? A bag contained 338 beads. Three children shared t
Sedaia [141]
B. 330/3
Each kid would get 112 beads without rounding the amount of beads

4 0
3 years ago
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Alex buys 2 1/2 pounds of grapes. He buys 1 1/4 times as many pounds of apples as grapes. How many pounds of apples does he buy?
galina1969 [7]
Given:
grapes = 2 1/2 pounds
apples = 1 1/4 times as many pounds as grapes.

For easier computation, let us convert these fractions into numbers with decimal places.

2 1/2 pounds = 2.50 pounds
1 1/4 times = 1.25

Apples = 2.50 pounds * 1.25 = 3.125 pounds of apples.

Or we still use fractions. We need to convert the mixed fractions into improper fractions.

2 1/2 = 5/2
1 1/4 = 5/4

5/2 * 5/4 = 5*5 / 2*4 = 25/8 is simplified to 3 1/8 pounds
3 0
3 years ago
Consider the first five steps of the derivation of the Quadratic Formula.
lara31 [8.8K]

Answer:

Full proof below

Step-by-step explanation:

\displaystyle ax^2+bx+c=0\\\\ax^2+bx=-c\\\\x^2+\biggr(\frac{b}{a}\biggr)x=-\frac{c}{a}\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\biggr(\frac{b}{2a}\biggr)^2=-\frac{c}{a}+\biggr(\frac{b}{2a}\biggr)^2\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=\frac{b^2-4ac}{4a^2}\\ \\x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\\ \\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

6 0
1 year ago
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