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3 years ago

Write in Expanded form: 75,300,207

Mathematics

2 answers:

Lubov Fominskaja [6]3 years ago

4 0

Answer:

$\huge\boxed{Answer\hookleftarrow}$

$75, 300, 207$

<h3>⎆ <u>In expanded form :-</u></h3>

$70,000,000 + 5,000,000 + 300,000 + 0 + 0 + 200 + 0 + 7$

# ꧁❣ RainbowSalt2²2² ࿐

aksik [14]3 years ago

3 0

70,000,000+5,000,000+300,000+200+7

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100 points<br> 10000000000000

Daniel [21]

Answer:

I’m like so broke

Step-by-step explanation:

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3 years ago

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Use the grouping method to factor this polynomial completely.<br> 3x3 + 12x2 + 2x + 8

dezoksy [38]

Answer:

(3x^2 + 2)(x + 4).

Step-by-step explanation:

3x3 + 12x2 + 2x + 8

3x^2(x + 4) + 2(x + 4) The x + 4 is common to the 2 groups so we have:

(3x^2 + 2)(x + 4).

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3 years ago

Lily hikes at least 1 hour but not more than 3 hours. She hikes at an average rate of 2.2 miles per hour. The distance Lily hike

Basile [38]

Average rate = distance / time
distance = average rate x time

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Required range is all real numbers from 2.2 to 6.6, inclusive.

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3 years ago

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Question

meriva

Question:

In a neighbourhood pet show, each of the animals entered is equally likely to win. if there are 7 dogs, 6 cats, 3 birds, and 2 gerbils entered, what is the probability that a bird will win the top prize?

Answer:

Probability that a bird will win the top prize is 0.167

Step-by-step explanation:

Given:

The number of dogs = 7

The number of cats = 6

The number of birds = 3

The number of gerbils = 2

To Find:

Probability that a bird will win the top prize = ?

Solution:

Let us first find the total number of pets .

The Total number of pets = 7 + 6 + 3 + 2 = 18

Now the probability of a bird will win the top prize is

=> $\frac{\text {Number of birds}}{\text{ Total number of pets}}$

=> $\frac{3}{18}$

=> $\frac{1}{6}$

=>0.167

4 0

3 years ago

Help please i don’t know how to do it

Norma-Jean [14]

Answer:

Step-by-step explanation:

$\frac{dy}{dx}=x^{2}(y-1)\\\frac{1}{y-1} \text{ } dy=x^{2} \text{ } dx\\\int \frac{1}{y-1} \text{ } dy=\int x^{2} \text{ } dx\\\ln|y-1|=\frac{x^{3}}{3}+C\\$

From the initial condition,

$\ln|3-1|=\frac{0^{3}}{3}+C\\\ln 2=C$

So we have that $\ln |y-1|=\frac{x^{3}}{3}+\ln 2\\e^{\frac{x^{3}}{3}+\ln 2}=y-1\\2e^{\frac{x^{3}}{3}}=y-1\\y=2e^{\frac{x^{3}}{3}}+1$

4 0

3 years ago