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3 years ago

Rounded 30,000 to smallest and the largest n

Mathematics

1 answer:

Elenna [48]3 years ago

6 0

3,000 is the smallest number and 300,000 is the largest number

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The average depth of the ocean is about 3.8 km.width of 7600 km. calculate the ratio depth:width?

Gnoma [55]

A ratio is like a fraction, so since you are provided with the numbers its easy to put it into a ratio
depth is 3.8km
width is 7600km
ratio is 3.8:7600

5 0

3 years ago

A cookie recipe calls for 3 cups of sugar and yields 24 cookies how much sugar will be needed to make 3 dozen cookies

Sindrei [870]

Answer:4.5 cups of sugar

Step-by-step explanation:

Okay so 24 cookies is 2 dozen

1 cup of sugar would be used for every 8 cookies

3 dozen is equal to 36 cookies

36/8=4.5 so you would need 4.5 cups of sugar

5 0

3 years ago

How can I solve these problems

KATRIN_1 [288]

Answer:

Step-by-step explanation:

Easy the three angles of a triangle equal 180 degrees

so 60+75 = 135 --> 180-135=

1) 145

2) 35

3) no since the tops a 100 and its a Isosceles triangle two corners left it be 40 for each corner

4) 34 for both x's cause 34+34+112. = 180

5) 180- 120= 60- 10 = 50/4 is (12.5)= x

6) x = 15 cause it be 120- 60 to get 60

4 0

3 years ago

Read 2 more answers

Please help me solve this problem

Stella [2.4K]

Answer:

See below

Step-by-step explanation:

<u>Since all graphs pass throught the origin, they all have the form of:</u>

y = ax², where a- coefficient

The coefficient of each graph's equation can be calculated based on the given pairs of points.

Each of them calculated for you and attached below.

8 0

3 years ago

F(x)=X over x^3-2x^2+5x why will this have no zeros

Mumz [18]

If you evaluate directly this function at x=0, you'll see that you have a zero denominator.

Nevertheless, the only way for a fraction to equal zero is to have a zero numerator, i.e.

$\dfrac{x}{x^3-2x^2+5x}=0\iff x=0$

So, this function can't have zeroes, because the only point that would annihilate the numerator would annihilate the denominator as well.

Moreover, we have

$\displaystyle \lim_{x\to 0} \dfrac{x}{x^3-2x^2+5x} = \lim_{x\to 0} \dfrac{x}{x(x^2-2x+5)} = \lim_{x\to 0} \dfrac{1}{x^2-2x+5} = \dfrac{1}{5}$

So, we can't even extend with continuity this function in such a way that $f(0)=0$

5 0

3 years ago