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kakasveta [241]

3 years ago

13

PLSSS HELP!! Don't just answer the questions for points, please. And don't say you can't see the picture.. if you can't see it d

on't answer :)

2 answers:

sweet-ann [11.9K]3 years ago

7 0

Answer:

It should be B

Step-by-step explanation:

pshichka [43]3 years ago

4 0

Answer:

Its B

Step-by-step explanation:

Hope this helps you,have an amazing day!!!!!!!!!!!!!!!

SRRY I MEANT TO PUT B.MY MISTAKE!!!!!! :))))))

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Dr dylan collects 2 hailstones during a storm. one hailstone weighs 2 3/8 pounds and the other hailstone weighs 1 3/10 pounds. h

kap26 [50]

Answer:

1 3/40 pounds

Step-by-step explanation:

difference in weight

= 2 3/8- 1 3/10

= 19/8-13/10

=(19*5 -13*4)/40

=95-52/40

=43/40

=1 3/40 pounds

8 0

3 years ago

Read 2 more answers

Please help me answer this question

stealth61 [152]

The value of dy/dx for the functions are

(i) $\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

<h3>Differentiation</h3>

From the question, we are to determine dy/dx for the given functions

(i) $x^{2} sin^{2}2x$

Let $u = x^{2}$

and

$v = sin^{2} 2x$

Also,

Let $w= sin2x$

∴ $v = w^{2}$

First, we will determine $\frac{dv}{dx}$

Using the Chain rule
$\frac{dv}{dx} = \frac{dv}{dw}.\frac{dw}{dx}$

$v = w^{2}$

∴ $\frac{dv}{dw} =2w$

Also,

$w= sin2x$

∴ $\frac{dw}{dx} =2cos2x$

Thus,

$\frac{dv}{dx} = 2w \times 2cos2x$

$\frac{dv}{dx} = 2sin2x \times 2cos2x$

$\frac{dv}{dx} = 4sin2x . cos2x$

Now, using the product rule

$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

From above

$u = x^{2}$

∴ $\frac{du}{dx}=2x$

Now,

$\frac{dy}{dx} = x^{2} (4sin2x.cos2x)+ sin^{2}2x (2x)$

$\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $xy^{2}+y^{2}x^{3} +2=0$

Then,

$x.2y\frac{dy}{dx}+ y^{2}(1)+y^{2}.3x^{2} + x^{3}.2y\frac{dy}{dx} +0=0$

$2xy\frac{dy}{dx}+ y^{2}+3x^{2}y^{2} + 2x^{3}y\frac{dy}{dx} =0$

$2xy\frac{dy}{dx}+2x^{3}y\frac{dy}{dx} =- y^{2}-3x^{2}y^{2}$

$\frac{dy}{dx} (2xy+2x^{3}y)=- y^{2}(1+3x^{2})$

$\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy+2x^{3}y}$

$\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy(1+x^{2}) }$

$\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

Hence, the value of dy/dx for the functions are

(i) $\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

Learn more on Differentiation here: brainly.com/question/24024883

#SPJ1

8 0

2 years ago

Evaluate 3 – 24 - 8 + 32

UNO [17]

Answer:

3

Step-by-step explanation:

Evaluate the calculation from left to right, that is

3 - 24 - 8 + 32

= - 21 - 8 + 32

= - 29 + 32

= 3

6 0

3 years ago

Read 2 more answers

What is the value of x? image below

prisoha [69]

Answer:

Angle x = 72

Step-by-step explanation:

Angle AHI = Angle HIK = 63

(This is because the two angles are corresponding angles, HI//JK)

Angle x = 180 - 63 - 45

= 72

3 0

2 years ago

Read 2 more answers

I will give you brainliest whoever answers correct first !

fomenos

Answer:

72.8

Step-by-step explanation:

16 is 8 times 2. 36.4 times 2 would be 72.8

8 0

3 years ago