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3 years ago

PLZ HELP!!! Will give brainiest if correct! Thxs!!!!!!!!

Mathematics

1 answer:

Mazyrski [523]3 years ago

4 0

$\frac{1.7 \times {10}^{ - 3} }{3 \times {10}^{2} } \\ = \frac{1.7}{3} \times {10}^{ - 3 - 2} \\ = 0.56 \times {10}^{ - 5} \\ = 5.6 \times {10}^{ - 6}$

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Triangle X Y Z is shown. Angle X Z Y is a right angle. Angle Z X Y is 60 degrees and angle X Y Z is 30 degrees. The length of hy

vladimir1956 [14]

Answer:

StartFraction StartRoot 3 EndRoot Over 3 EndFraction

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

we know that

In the right triangle XYZ

$cos(Y)=\frac{ZY}{XY}$ ---> adjacent side divided by the hypotenuse

substitute the values

$cos(30\°)=\frac{ZY}{4}$

Remember that

$cos(30\°)=\frac{\sqrt{3}}{2}$

substitute

$\frac{\sqrt{3}}{2}=\frac{ZY}{4}$

$ZY=(4)\frac{\sqrt{3}}{2}$

$ZY=2\sqrt{3}\ units$

step 2

$sin(Y)=\frac{XZ}{XY}$ --> opposite side divided by the hypotenuse

substitute the values

$sin(30\°)=\frac{XZ}{4}$

Remember that

$sin(30\°)=\frac{1}{2}$

$\frac{1}{2}=\frac{XZ}{4}$

$XZ=2\ units$

step 3

$tan(Y)=\frac{XZ}{ZY}$ --> opposite side divided by adjacent side

substitute the values

$tan(Y)=\frac{2}{2\sqrt{3}}$

$tan(Y)=\frac{1}{\sqrt{3}}$

Simplify

$tan(Y)=\frac{\sqrt{3}}{3}$

StartFraction StartRoot 3 EndRoot Over 3 EndFraction

7 0

3 years ago

Read 2 more answers

Please help answer this

yuradex [85]

I think you can use cosine rule if only the lines that form the <52 are equal.

6 0

3 years ago

What is the simplified form of √45<br>A) √5<br>B) √9•5<br>C) 3√5<br>D)5√3

Kaylis [27]

The simplified form is C .

4 0

3 years ago

Try to sketch by hand the curve of intersection of the parabolic cylinder y = x2 and the top half of the ellipsoid x2 + 7y2 + 7z

vovikov84 [41]

Plug $y=x^2$ into the equation of the ellipsoid:

$x^2+7(x^2)^2+7z^2=49$

Complete the square:

$7x^4+x^2=7\left(x^4+\dfrac{x^2}7+\dfrac1{14^2}-\dfrac1{14^2}\right)=7\left(x^2+\dfrac1{14}\right)^2+\dfrac1{28}$

Then the intersection is such that

$7\left(x^2+\dfrac1{14}\right)^2+7z^2=\dfrac{1371}{28}$

$\left(x^2+\dfrac1{14}\right)^2+z^2=\dfrac{1371}{196}$

which resembles the equation of a circle, and suggests a parameterization is polar-like coordinates. Let

$x(t)^2+\dfrac1{14}=\sqrt{\dfrac{1371}{196}}\cos t\implies x(t)=\pm\sqrt{\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}}$

$y(t)=x(t)^2=\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}$

$z=\sqrt{\dfrac{1371}{196}}\sin t$

(Attached is a plot of the two surfaces and the intersection; red for the positive root $x(t)$ , blue for the negative)

4 0

3 years ago

B 5 3 a 4 What is the length of the hypotenuse of this triangle? 3 units 5 units 4 units

Lunna [17]

Answer:

5 units

Step-by-step explanation:

The hypotenuse is always the longest segment on a triangle. If the lengths of the sides of this triangle are 3 units, 4 units, and 5 units, the hypotenuse will be 5 units. You can use the pythagorean theorem to check.

a^2+b^2=c^2

If you substitute the variable with the lengths of the legs in ascending order, you will get 3^2+4^2=5^2. Simplifying the equation, you will get 9+16=25, or 25=25. Thus, proving the equation.

Hope this helped! :)

6 0

3 years ago