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Harlamova29_29 [7]

3 years ago

8

I cant do this anymore

1 answer:

cricket20 [7]3 years ago

7 0

I can’t either Stay strong

You might be interested in

Simplify 2x(6x^3-4x+11)

ollegr [7]

<span><span> $12 x^{4} -8 x^{2} +2 2^{x}$ </span><span>

</span></span>

8 0

3 years ago

Simplify the following surds :<br><br> 1) 2√21 × √27 ÷ √343<br> 2) 7√5 × √125 ÷ 2√27

ludmilkaskok [199]

Answer:

1) $\frac{18}{7}$

2) $\frac{175\sqrt{3}}{18}$

Step-by-step explanation:

* Lets explain how to simplify a square root

1)

∵ $2\sqrt{21}$ × $\sqrt{27}$ ÷ $\sqrt{343}$

∵ $\sqrt{21}=\sqrt{3}$ × $\sqrt{7}$

∴ $2\sqrt{21}$ = $2\sqrt{3}$ × $\sqrt{7}$

∵ $\sqrt{27}$ = $\sqrt{3}$ × $\sqrt{3}$ × $\sqrt{3}$

∵ $\sqrt{3}$ × $\sqrt{3}$ = 3

∴ $\sqrt{27}$ = $3\sqrt{3}$

∴ $2\sqrt{21}$ × $\sqrt{27}$ =

$2\sqrt{3}$ × $\sqrt{7}$ × $3\sqrt{3}$

∵ $\sqrt{3}$ × $\sqrt{3}$ = 3

∵ 2 × 3 × 3 = 18

∴ $2\sqrt{21}$ × $\sqrt{27}$ = $18\sqrt{7}$

∵ $\sqrt{343}$ = $\sqrt{7}$ × $\sqrt{7}$ × $\sqrt{7}$

∵ $\sqrt{7}$ × $\sqrt{7}$ = 7

∴ $\sqrt{343}$ = $7\sqrt{7}$

∵ $2\sqrt{21}$ × $\sqrt{27}$ ÷ $\sqrt{343}$ =

$18\sqrt{7}$ ÷ $7\sqrt{7}$

∵ $\sqrt{7}$ ÷ $\sqrt{7}$ = 1

∴ $2\sqrt{21}$ × $\sqrt{27}$ ÷ $\sqrt{343}$ =

$\frac{18}{7}$

2)

∵ $7\sqrt{5}$ × $\sqrt{125}$ ÷ $2\sqrt{27}$

∵ $\sqrt{125}$ = $\sqrt{5}$ × $\sqrt{5}$ × $\sqrt{5}$

∵ $\sqrt{5}$ × $\sqrt{5}$ = 5

∴ $\sqrt{125}$ = $5\sqrt{5}$

∴ $7\sqrt{5}$ × $\sqrt{125}$ =

$7\sqrt{5}$ × $5\sqrt{5}$

∵ $\sqrt{5}$ × $\sqrt{5}$ = 5

∴ $7\sqrt{5}$ × $\sqrt{125}$ = 7 × 5 × 5 = 175

∵ $2\sqrt{27}$ = $2\sqrt{3}$ × $\sqrt{3}$ × $\sqrt{3}$

∵ $\sqrt{3}$ × $\sqrt{3}$ = 3

∴ $2\sqrt{27}$ = $6\sqrt{3}$

∴ $7\sqrt{5}$ × $\sqrt{125}$ ÷ $2\sqrt{27}$ =

175 ÷ $6\sqrt{3}$ = $\frac{175}{6\sqrt{3}}$

∵ $\frac{175}{6\sqrt{3}}$ not in the simplest form because

the denominator has square root

∴ Multiply up and down by $\sqrt{3}$

∴ $\frac{175}{6\sqrt{3}}$ = $\frac{175\sqrt{3}}{6\sqrt{3}*\sqrt{3}}$

∴ $\frac{175}{6\sqrt{3}}$ = $\frac{175\sqrt{3}}{18}$

∴ $7\sqrt{5}$ × $\sqrt{125}$ ÷ $2\sqrt{27}$ =

$\frac{175\sqrt{3}}{18}$

7 0

3 years ago

If w = 5 cos (xy) − sin (xz) and x = 1/t , y = t, z = t^3 ; then find dw/dt

Scrat [10]

In this question, we find the derivatives, using the chain's rule.

Doing this, the derivative is:

$\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}$

Chain Rule:

Suppose we have a function $w(x,y,z), x = x(t), y = y(t), z = z(t)$ , and want to find it's derivative as function of t. It will be given by:

$\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}$

Thus, we have to find the desired derivatives, which are:

w of x:

$\frac{dw}{dx} = -5y\sin{(xy)} - z\cos{(xz)}$

Considering $x = \frac{1}{t}, y = t, z = t^3$

$\frac{dw}{dx} = -5t\sin{(1)} - t^3\cos{(t^2)}$

w of y:

$\frac{dw}{dy} = -5x\cos{(xy)}$

Considering $x = \frac{1}{t}, y = t$

$\frac{dw}{dy} = -\frac{5}{t}\cos{1}$

w of z:

$\frac{dw}{dz} = -x\cos{(xz)}$

Considering $x = \frac{1}{t}, z = t^3$

$\frac{dw}{dz} = -\frac{1}{t}\cos{(t^2)}$

Derivatives of x, y and z as functions of t:

$\frac{dx}{dt} = -\frac{1}{t^2}$

$\frac{dy}{dt} = 1$

$\frac{dz}{dt} = 3t^2$

Derivative of w as function of t.

Now, we just replace what we found into the formula. So

$\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}$

$\frac{dw}{dt} = (-5t\sin{(1)} - t^3\cos{(t^2)})(-\frac{1}{t^2}) - (\frac{5}{t}\cos{1}) - (\frac{1}{t}\cos{(t^2)})3t^2$

Applying the multiplications:

$\frac{dw}{dt} = \frac{5}{t}\sin{1} + t\cos{t^2} - \frac{5}{t}\cos{1} - 3t\cos{t^2}$

Applying the simplifications:

$\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}$

Which is the derivative.

For more on the chain rule, you can check brainly.com/question/12795383

8 0

3 years ago

Multiply the expression

user100 [1]

Answer:

-26.

Step-by-step explanation:

4 * 2 * (-3) + (-10) / 5

Deal with the parentheses first (PEMDAS).

= 4 * 2 * -3 - 10 / 5

Now the multiplications and divisions:

= 8 * -3 - 2

= -24 - 2

= -26.

8 0

3 years ago

Can someone help me on these questions?

ch4aika [34]

Me not good on math matics I sorry hybrid

6 0

3 years ago

Read 2 more answers