Please help me!! Find the area of each triangle (A-D). Show your reasoning.

Use the following set of data: 78, 78, 83, 85, 89, 91, 95, 98 What is the range for the set of data? 78 87 20 98

QveST [7]

Hello!

The range is the difference between the largest and smallest numbers in a data set.

98-78=20

The range is 20.

I hope this helps!

4 0

3 years ago

Pls solve this question

just olya [345]

Answer:

${( \sqrt{ {x}^{ - 3} }) }^{5} = ({( {x}^{ - 3}) }^{ \frac{1}{2} } ) ^{5} \\ \\ = {x}^{ - 3 \times \frac{1}{2} \times 5} = {x}^{ - \frac{15}{2} } = \frac{1}{ {x}^{ \frac{15}{2} } }$

Or ;

${( \sqrt{ {x}^{ - 3} } )}^{5} = {( \sqrt{ \frac{1}{ {x}^{3} }} })^{5} = ( { \frac{ \sqrt{1} }{ \sqrt{ {x}^{3} } } })^{5} \\ \\ = ( { \frac{1}{ \sqrt{x} \sqrt{ {x}^{2} } } })^{5} = ({ \frac{1}{ x\sqrt{x} }})^{5} \\ \\ = ( \frac{ {(1)}^{5} }{ {x}^{5} ( \sqrt{x} )^{5} } ) = \frac{1}{ {x}^{5} \times {x}^{ \frac{1}{2} \times 5 } } \\ \\ = \frac{1}{ {x}^{5} \times {x}^{ \frac{5}{2} } } = \frac{1}{ {x}^{5} \sqrt{ {x}^{5} } } \\ \\ = \frac{1}{ {x}^{5} \sqrt{ {x}^{4} {x}^{1} } } = \frac{1}{ {x}^{5} \sqrt{x} \sqrt{( { {x}^{2} })^{2} } } \\ \\ = \frac{1}{ {x}^{5} {x}^{2} \sqrt{x} } = \frac{1}{ {x}^{7} \sqrt{x} } = \frac{ \sqrt{x} }{ {x}^{8} }$

3 0

2 years ago

Dylan has $14.00 to spend at the book fair. He decides to buy two paperbacks for $4.50 each. Then, he sees some sticker books. A

KonstantinChe [14]

Answer: 5 dollars

Step-by-step explanation:

7 0

3 years ago

Find the intervals on which f(x) is increasing, the intervals on which f(x) is decreasing, and the local extrema. f (x )equa

Kaylis [27]

Answer:

f(x) is increasing in the intervals (-∞,-4) and (9,∞)

f(x) is decreasing in the ingercal (-4,9)

The local extrema is:

local max at (-4,496)

and local min at (9,-1701)

Step-by-step explanation:

In order to solve this problem we must start by finding the derivative of the provided function, which we can find by using the power rule.

if $f(x)=ax^{n}$ then $f'(x)=anx^{n-1}$

so we get:

$f(x)=2x^{3}-15x^{2}-216x$

$f'(x)=6x^{2}-30x-216$

in order to find the critical points we mus set the derivative equal to zero, since the local max an min will happen when the slope of the tangent line to the given point is zero, so we get:

$6x^{2}-30x-216=0$

we can solve this by factoring, so let's factor that equation:

6(x+4)(x-9)=0

we can now set each of the factors equal to zero so we get:

x+4=0 and x-9=0

when solving each for x we get that:

x=-4 and x=9

These are our critical points, now we can build the possible intervals we are going to use to determine where the function will be increasing and where it will be decreasing:

(-∞,-4), (-4,9) and (9,∞)

so now we need to test these intervals in the derivative to see if the graph will be increasing or decreasing in the given intervals. So let's pick x=-5 for the first one, x=0 for the second one and x=10 for the third one.

When evaluating them into the first derivative we get that:

f'(-5)=84, this is a positive answer so it means that the function is increasing in the interval (-∞,--4)

f'(0)=-216, this is a negative anser so it means that the function is decreasing in the interval (-4,9)

f'(10)=84, this is a positive answer so it means that the function is increasing in the interval (9,∞)

Now, for the local extrema, we can see that at x=-4, the function it's increasing on the left of this point while it's decreasing to the right, which means that there will be a local maximum at x=-4, so the local max is the point (-4,496)

We can see that at x=9, the function it's decreasing on the left of this point while it's increasing to the right, which means that there will be a local minimum at x=-9, so the local min is the point (9,-1701)

7 0

4 years ago

How to translate f(x)(blue) to g(x) red?

Wittaler [7]

Answer:

g(x) = $\sqrt{x+2}$ - 3

Step-by-step explanation:

g(x) = $\sqrt{x+2}$ - 3

7 0

3 years ago