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xxMikexx [17]
3 years ago
15

I need help with this question "To estimate the product 3.48 x 7.33,Marisa multiplied 4x8 to get 32.Explain how she can make a c

loser estimate."
Mathematics
1 answer:
Furkat [3]3 years ago
3 0
3.5 x 7.3
Take it to the nearest tenth.
Does that answer your question?
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A runner finished a total of 1500 kilometers in races before retiring she finished thirty-two 15 kilometer races. The rest of he
Triss [41]

Answer:

102 of runner's races were 10-kilometer races.

8 0
2 years ago
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Hiiiooo! Can someone please help ❤️
V125BC [204]

Answer:

B. √21

Step-by-step explanation:

x^2 = √5 x √5 + 4 x 4

x^2 = 5 + 16

x^2 = 21

x = √21

<em>Hope that helps! :)</em>

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<em>-Aphrodite</em>

4 0
3 years ago
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Divide. Write the quotient in lowest terms.<br> 5 divide 3 1/3
Step2247 [10]

Answer:

3/2 or 1 1/2

Step-by-step explanation:

5÷3 1/3

5/1 ÷ 10/3 because you have to get it into an improper fraction form in order to divide

5/1 × 3/10 because you have to multiply the first number by the reciprocal

5  and 10 cancel out so you would have 2 at the bottom

3/2 or 1 1/2

3 0
3 years ago
Write an inequality for x that would give this rectangle a perimeter of at least 300ft
Travka [436]

Answer:

The inequality for x is:

x\geq 142

Step-by-step explanation:

Given:

Width of rectangle = 3 ft

Height or length of rectangle = (x+5) ft

Perimeter is at least 300 ft

To write an inequality for x.

Solution:

Perimeter of a rectangle is given as:

⇒ 2l+2w

where l represents length of the rectangle and w represents the width of the rectangle.

Plugging in the given values in the formula, the perimeter can be given as:

⇒ 2(x+5)+2(3)

Using distribution:

⇒ 2x+10+6

Simplifying.

⇒ 2x+16

The perimeter is at lest 300 ft. So, the inequality can be given as:

⇒ 2x+16\geq 300

Solving for x

Subtracting both sides by 16.

⇒ 2x+16-16\geq 300-16

⇒ 2x\geq284

Dividing both sides by 2.

⇒ \frac{2x}{2}\geq \frac{284}{2}

⇒ x\geq 142   (Answer)

7 0
3 years ago
How do you find the limit?
coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}

We can cancel x-5.

\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}

Then directly substitute x = 5 in.

\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\&#10;&#10;\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\&#10;&#10;\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s <em>too easy</em> but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}

Differentiate numerator and denominator, apply power rules.

<u>Differential</u> (Power Rules)

\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}

<u>Differentiation</u> (Property of Addition/Subtraction)

\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}

Hence from the expressions,

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}

<u>Differential</u> (Constant)

\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \  is \ \ a \ \ constant.)}}

Therefore,

\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}

Now we can substitute x = 5 in.

\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}

Thus, the limit value is 2/5 same as the first method.

Notes:

  • If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.
8 0
3 years ago
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