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3 years ago

WILL GIVE BRAINLIEST NO LINKS

Mathematics

1 answer:

Marrrta [24]3 years ago

8 0

Answer:

https://www.calculatorsoup.com/ go to calculator soup! it's literally just a bunch of calculators, i'm sure you can find one!

You might be interested in

The ratio of the surface areas of two similar solids is 49:100. What is the ratio of their corresponding side lengths?

algol13

<span>ratio of surface area = 49:100
ratio of corresponding lengths = √49:√100 = 7:10
answer: C

Hope this helps

</span>

3 0

3 years ago

Read 2 more answers

A researcher says to the respondents in a poll, “Eating too many sugary foods leads to cavities. Would you rather have soda or w

photoshop1234 [79]

Answer:

. D. Yes, the more information provided by a researcher the better. Respondents can now give an informed opinion and the results will be more accurate.

Step-by-step explanation:

But again this could be an opinion answer as well

Hope this helps

If this seems incorrect anyway please just comment and I shall change my answer thanks very much :)

6 0

3 years ago

I have included a picture of the equation <br><br>

Alchen [17]

Answer:

it's c , x= 1.5 or -4

2x^2 + 5x - 12 = 0

First factor the left side of the equation

(2x-3)(x+4)=0

Second, set the factors equal to 0

2x-3=0 or x+4=0

+3 +3 -4 -4

2x=3 -4

divide both side by 2 and you get 3/2 which equals 1.5

x=1.5 or x= -4

3 0

2 years ago

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 6x(1/3) + 3x(4/3). You must justi

stealth61 [152]

Applying our power rule gets us our first derivative,

$\rm f'(x)=6\frac13x^{-2/3}+3\cdot\frac43x^{1/3}$

simplifying a little bit,

$\rm f'(x)=2x^{-2/3}+4x^{1/3}$

looking for critical points,

$\rm 0=2x^{-2/3}+4x^{1/3}$

We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.

$0=2x^{-2/3}\left(1+2x\right)$

When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.

Then apply your Zero-Factor Property,

$\rm 0=2x^{-2/3}\qquad\qquad\qquad 0=(1+2x)$

and solve for x in each case to find your critical points.

Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.

Let's come back to this,

$\rm f'(x)=2x^{-2/3}+4x^{1/3}$

and take our second derivative.

$\rm f''(x)=-\frac43x^{-5/3}+\frac43x^{-2/3}$

Looking for inflection points,

$\rm 0=-\frac43x^{-5/3}+\frac43x^{-2/3}$

Again, pulling out the smaller power of x, and fractional part,

$\rm 0=-\frac43x^{-5/3}\left(1-x\right)$

And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.

Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.

8 0

3 years ago

What is the trigonometric ratio for cos D ?

leva [86]

Answer:

Here,

H=75
P=72
B=21

We know that cosD= b/h

=21/75

=7/25

7 0

2 years ago

WILL GIVE BRAINLIEST *NO LINKS*

WILL GIVE BRAINLIEST NO LINKS