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2 years ago

In a few sentences, describe the role that the ovaries, fallopian tubes, and uterus play in the female reproductive system.

2 answers:

7 0

Ovaries: produce egg or ovum and hormones

Fallopian tubes: transport egg to uterus

Uterus: to accept the fertilized egg and nourishes it.

Dmitry [639]2 years ago

7 0

The ovaries produce the female egg cells, called the ova or oocytes. The oocytes are transported through the fallopian tubes and there is where Conception, the fertilization of an egg by a sperm, happens. The fertilized egg then moves to the uterus. Because of the hormones of the reproductive cycle the uterine lining has thickened and the fertilized egg can implant into thickened uterine lining and continue to develop.

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vesna_86 [32]

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Explanation:

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2 years ago

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3 years ago

If 2^x=3^y=12^z, show that 1÷z=1÷y+2÷x.....

Ostrovityanka [42]

Answer:

Proof of $\dfrac{1}{z}}=\dfrac{1}{y}+\dfrac{2}{x}$ is shown below.

Explanation:

The given equation is

$2^x=3^y=12^z$

To prove : $\dfrac{1}{z}=\dfrac{1}{y}+\dfrac{2}{x}$

Proof : Let as assume,

$2^x=3^y=12^z=k$

$2^x=k\Rightarrow 2=k^{\frac{1}{x}}$ ...(1)

$3^y=k\Rightarrow 3=k^{\frac{1}{y}}$ ...(2)

$12^z=k\Rightarrow 12=k^{\frac{1}{z}}$ ...(3)

We know that

$12=2\times 2\times 3$

$12=(2)^2\times 3$

Using (1), (2) and (3), we get

$k^{\frac{1}{z}}=(k^{\frac{1}{x}})^2\times k^{\frac{1}{y}}$

$k^{\frac{1}{z}}=k^{\frac{2}{x}}\times k^{\frac{1}{y}}$

$k^{\frac{1}{z}}=k^{\frac{2}{x}+\frac{1}{y}}$

On comparing both sides, we get

$\dfrac{1}{z}}=\dfrac{2}{x}+\dfrac{1}{y}$

$\dfrac{1}{z}}=\dfrac{1}{y}+\dfrac{2}{x}$

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3 years ago

Which of the following BEST describes a body in equilibrium?

sukhopar [10]

$\textsf {\large{\underline { \underline {SOLUTION:-}}}}$

⇰Let O be the point of observation on the ground OX.

⇰Let A and B be the two positions of the jet.

$\bf➠Then, \angle XOA=60⁰ \: and \: \angle \: XOB=30⁰ \\ \\$

$\bf ➠Draw \: AL \perp \: OX \: and \: BM \perp \: OX \\ \\$

$\bf \:⇰ Let \: AL=BM=h \: meters. \\ \\$

$\bf \: ⇰ \: Speed \: of \: Jet = 720 \: km/hr\\$

$\bf = \huge( \small \: 720 \times \frac{5}{18} \huge) \small \: m / s\\$

$\bf = 200 \: m/s$

$\bf \:➥ Time \: taken \: to \: cover \: the \: distance \: AB=15 \: sec \\ \\$

$\bf➥ Distance \: covered =(speed×time) \\ \\$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = (200 \times 15)m = 3000m \\ \\$

$\bf \therefore LM=AB=3000m \\ \\$

${ \bf \: ⇰ \: Let \: OL} = x \: \bf m \\ \\$

$\bf➥From \: right \: \triangle OLA, \: we \: have: \\$

$\bf \frac{OL}{AL} =cot \: 60⁰= \frac{1}{ \sqrt{3} } \implies \frac{x}{h} = \frac{1}{ \sqrt{3} } \\ \\$

$\bf \implies \: x = \frac{h}{ \sqrt{3} } \: \: \: \: \: \: \: ....(1) \\ \\$

$\bf \:➥ From \: right \: \triangle \: OMB,we \: have: \\$

$\bf\frac{OM}{BM} =cot30⁰= \sqrt{3} \\ \\$

$\bf \implies \: \frac{x + 3000}{h} = \sqrt{3} \\ \\$

$\bf[ \boxed{➯OM=OL+LM=OL+AB=(x+3000)m \: and \: BM=h \: m]}$

$\bf \implies \: x + 3000 = \sqrt{3} h \\$

$\bf \implies \: x = ( \sqrt{3} h - 3000) \: \: \: \: \: \: ....(2) \\$

⇰ Equating the value of x from (1) and (2),we get;

$\bf \: \sqrt{3} h - 3000 = \frac{h}{ \sqrt{3} } \\ \\$

$\implies \bf \: 3h - 3000 \sqrt{3} = h \\$

$\bf \implies \: 2h = (3000 \times \sqrt{3} ) = (3000 \times 1.762) \\$

$\bf \implies \: h = (3000 \times 0.866) = 2598 \\ \\$

$\bf \:➥ Hence \: ,the \: required \: height \: is \: \boxed{ \boxed{ \bf2598 \: m.}} \\ \\$

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4 0

2 years ago

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