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3 years ago

4 over 5 divided by 15

Mathematics

2 answers:

kirill [66]3 years ago

5 0

<h3>Answer</h3>

$\tt \large \: \purple ⇝ \frac{4}{5} \div 15$

$\: \:$

$\tt \large \: \purple ⇝ = \frac{4}{5} \times \frac{1}{15}$

$\: \:$

$\tt \large \: \purple ⇝= \frac{4}{75}$

$\: \:$

<h3>Hope Helps</h3>

Alex Ar [27]3 years ago

3 0

Step-by-step explanation:

4/5 ÷ 15

= 4 / 5 × 1/15

=4/75

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3 years ago

Consider the following function. f(x) = 16 − x2/3 Find f(−64) and f(64). f(−64) = f(64) = Find all values c in (−64, 64) such th

VARVARA [1.3K]

Answer:

This does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64).

Step-by-step explanation:

The given function is

$f(x)=16-\frac{x^2}{3}$

To find $f(-64)$ , we substitute $x=-64$ into the function.

$f(-64)=16-\frac{(-64)^2}{3}$

$f(-64)=16-\frac{4096}{3}$

$f(-64)=-\frac{4048}{3}$

To find $f(64)$ , we substitute $x=64$ into the function.

$f(64)=16-\frac{(64)^2}{3}$

$f(64)=16-\frac{4096}{3}$

$f(64)=-\frac{4048}{3}$

To find $f'(c)$ , we must first find $f'(x)$ .

$f'(x)=-\frac{2x}{3}$

This implies that;

$f'(c)=-\frac{2c}{3}$

$f'(c)=0$

$\Rightarrow -\frac{2c}{3}=0$

$\Rightarrow -\frac{2c}{3}\times -\frac{3}{2}=0\times -\frac{3}{2}$

$c=0$

For this function to satisfy the Rolle's Theorem;

It must be continuous on $[-64,64]$ .

It must be differentiable on $(-64,64)$ .

and

$f(-64)=f(64)$ .

All the hypotheses are met, hence this does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64) is the correct choice.

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3 years ago

Read 2 more answers

I rlly need help pls help me

kifflom [539]

Answer:

This one ↓↓↓↓↓↓

Step-by-step explanation:

The picture attached is the reflection over the line y = 0

-Chetan K

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2 years ago

How would I figure out a linear equation with two plot points with fractions? (1/3(6),3) and (1/2(19),-2) are my two plot points

murzikaleks [220]

Answer: $m=\frac{-30}{79},\ y=5.40$

Step-by-step explanation:

Given

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Convert mixed numeral to fraction

$\left(6\frac{1}{3},3\right)\Rightarrow \left(\dfrac{19}{3},3\right)\\\\\left(19\frac{1}{2},-2\right)\Rightarrow \left(\dfrac{39}{2},-2\right)$

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$\Rightarrow m=\dfrac{3-(-2)}{\dfrac{19}{3}-\dfrac{39}{2}}\\\\\Rightarrow m=\dfrac{5}{\dfrac{38-117}{6}}\\\\\Rightarrow m=\dfrac{-30}{79}$

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3 years ago

What is the answer for part b

trasher [3.6K]

Answer:

(-3,-10)

Step-by-step explanation:

Given part a, you seem to know the midpoint formula. We can easily edit that to figure out what an endpoint is. Say that a variable "a" represents the x coordinate of D, and b the y coordinate. As the x coordinate of M is 0.5, and the x coordinate of C is 2, we can use the midpoint formula to say that $\frac{a+2}{2} =-0.5$ . Multiplying both side by 2, we get that a+2=-1, and subtracting both sides by 2, a=-3. Similarly, $\frac{b+7}{2} =-1.5$ , b+7=-3, and b=-10, making our answer (-3, -10). Let me know if you have any questions.

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3 years ago

4 over 5 divided by 15​

4 over 5 divided by 15