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3 years ago

Solve the equation for z, where z is restricted to the given interval.

1 answer:

7 0

$y = 7sin(z)\implies 0 = 7sin(z)\implies 0=sin(z)\implies sin^{-1}(0)=sin^{-1}[sin(z)] \\\\\\ sin^{-1}(0)=z\implies 0=z$

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Lara found the volume of a sphere with a diameter of 10.

Paraphin [41]

Answer:

No, she used the diameter instead of the radius measure.

Step-by-step explanation:

we know that

The volume of the sphere is equal to

$V=\frac{4}{3}\pi r^{3}$

we have

$r=10/2=5\ units$ ----> the radius is half the diameter

substitute

$V=\frac{4}{3}\pi (5)^{3}$

$V=\frac{500}{3}\pi\ units^{3}$

3 0

3 years ago

Read 2 more answers

What is 3 divided 162

dmitriy555 [2]

Answer:

0.185185185185185185.........

Step-by-step explanation:

i used a calculator, to the nearst tenth is 0.18 to the nearest 100th is 0.185 also the 185 is repeating so u put a line over the numbers 185

4 0

3 years ago

Read 2 more answers

Please help me with this math!

wel

Answer:

Please check the explanation.

Step-by-step explanation:

Given the polynomial function

$f\left(x\right)\:=x^3+2x^2-16x-32$

Let us determine the factors by solving

$\:\left0\right\:=x^3+2x^2-16x-32\:$

$\left(x+2\right)\left(x+4\right)\left(x-4\right)=0$

Using the zero factor principle:

if $ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$x+2=0\quad \mathrm{or}\quad \:x+4=0\quad \mathrm{or}\quad \:x-4=0$

Thus, (x+2) (x+4) and (x-4) are the factors of the polynomial function.

Therefore,

YES

(x+2)

(x+4)

(x-4)

And (x-2) (x+6) are not the factors of the polynomial function.

Therefore,

YES

(x-2)
(x+6)

3 0

3 years ago

OK so I was going through work and I come across this question 3/4 x 4/7 do you know it?

mezya [45]

Yes i do, the answer is 0.42

6 0

3 years ago

Simplify: [5×(25)^n+1 - 25 × (5)^2n]/[5×(5)^2n+3 - (25)^n+1]

Alekssandra [29.7K]

$\green{\large\underline{\sf{Solution-}}}$

Given expression is 

$\rm :\longmapsto\:\dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} }$

can be rewritten as

$\rm \: = \: \dfrac{5 \times { {(5}^{2} )}^{n + 1} - {5}^{2} \times {5}^{2n} }{5 \times {5}^{2n + 3} - {( {5}^{2} )}^{n + 1} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {( {x}^{m} )}^{n} \: = \: {x}^{mn}}}} \\$

And

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}} \\$

So, using this identity, we

$\rm \: = \: \dfrac{5 \times {5}^{2n + 2} - {5}^{2n + 2} }{{5}^{2n + 3 + 1} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 4} - {5}^{2n + 2} }$

can be further rewritten as

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 2 + 2} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2} - 1)}$

$\rm \: = \: \dfrac{4}{25 - 1}$

$\rm \: = \: \dfrac{4}{24}$

$\rm \: = \: \dfrac{1}{6}$

Hence, 

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} } = \frac{1}{6} }}$

4 0

3 years ago

Solve the equation for​ z, where z is restricted to the given interval.

Solve the equation for z, where z is restricted to the given interval.