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2 years ago

What is the measure of angle Y? 60 40

Mathematics

1 answer:

Thepotemich [5.8K]2 years ago

4 0

Answer:

if it is triangle it is 80

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What is the answer this equation 3/8 × 32/5

cupoosta [38]

96/40
simplified 12/5

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3 years ago

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Find the volume of the parallelepiped with adjacent edges t = 3j – 4j – 6k, u = 3i – 7j – 7k and v = 5i + j + 3k.

Assoli18 [71]

Answer:

94 cubic units

Step-by-step explanation:

The volume of this parallelepiped (the product of area of its base and height) is equal to the absolute value of the scalar triple product $|\vec{t}\cdot (\vec{u}\times \vec{v})|$

Since

$\vec{t}=3\vec{i}-4\vec{j}-6\vec{k}=(3,-4,-6)\\ \\\vec{u}=3\vec{i}-7\vec{j}-7\vec{k}=(3,-7,-7)\\ \\\vec{v}=5\vec{i}+\vec{j}+3\vec{k}=(5,1,3)$

we have

$|\vec{t}\cdot (\vec{u}\times \vec{v})|=\left|\left|\begin{array}{ccc}3&-4&-6\\3&-7&-7\\5&1&3\end{array}\right|\right|=|3\cdot (-7)\cdot 3+(-4)\cdot (-7)\cdot 5+3\cdot 1\cdot (-6)-5\cdot (-7)\cdot (-6)-3\cdot (-7)\cdot 1-3\cdot (-4)\cdot 3|=|-63+140-18-210+21+36|=|-94|=94\ un^3.$

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2 years ago

Which is the volume of a cube if each side measures 1 foot

grandymaker [24]

volume of a cube
V= a^3 = 1^3 = 1

answer

D . 1 cubic foot

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3 years ago

Find the volume of the cone please help me

Ostrovityanka [42]

Here,

Height (h) = 5

Radius (r) = 5

Volume = ?

Now,

◆ Volume = 1/3πr²h

= 1/3*3.14*5*5*5

= 130.83 unit^3

I hope you understand...

It is Right answer...

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5 0

2 years ago

100 points! simplify write as a product compute

Rom4ik [11]

Answer:

a) $\sqrt{61 - 24 \sqrt{5} } = - 4 + 3 \sqrt{5}$

b) $( \sqrt{ ( {c}^{2} - 1) ({b}^{2} - 1) } - {2 \sqrt{bc} }) (\sqrt{ ( {c}^{2} - 1) ({b}^{2} - 1) } + {2 \sqrt{bc} } )$

c) $\frac{ \sqrt{9 - 4 \sqrt{5} } }{2 - \sqrt{5} } = - 1$

Step-by-step explanation:

We want to simplify

$\sqrt{61 - 24 \sqrt{5} }$

Let :

$\sqrt{61 - 24 \sqrt{5} } = a - b \sqrt{5}$

Square both sides of the equation:

$(\sqrt{61 - 24 \sqrt{5} } )^{2} = ({a - b \sqrt{5} })^{2}$

Expand the RHS;

$61 - 24 \sqrt{5} = {a}^{2} - 2ab \sqrt{5} + 5 {b}^{2}$

Compare coefficients on both sides:

${a}^{2} + 5 {b}^{2} = 61 - - - (1)$

$- 24 = - 2ab \\ ab = 12 \\ b = \frac{12}{b} - - -( 2)$

Solve the equations simultaneously,

$\frac{144}{ {b}^{2} } + 5 {b}^{2} = 61$

$5 {b}^{4} - 61 {b}^{2} + 144 = 0$

Solve the quadratic equation in b²

${b}^{2} = 9 \: or \: {b}^{2} = \frac{16}{5}$

This implies that:

$b = \pm3 \: or \: b = \pm \frac{4 \sqrt{5} }{5}$

When b=-3,

$a = - 4$

Therefore

$\sqrt{61 - 24 \sqrt{5} } = - 4 + 3 \sqrt{5}$

We want to rewrite as a product:

${b}^{2} {c}^{2} - 4bc - {b}^{2} - {c}^{2} + 1$

as a product:

We rearrange to get:

${b}^{2} {c}^{2} - {b}^{2} - {c}^{2} + 1- 4bc$

We factor to get:

${b}^{2} ( {c}^{2} - 1) - ({c}^{2} - 1)- 4bc$

Factor again to get;

$( {c}^{2} - 1) ({b}^{2} - 1)- 4bc$

We rewrite as difference of two squares:

$(\sqrt{( {c}^{2} - 1) ({b}^{2} - 1) })^{2} - ( {2 \sqrt{bc} })^{2}$

We factor the difference of square further to get;

$( \sqrt{ ( {c}^{2} - 1) ({b}^{2} - 1) } - {2 \sqrt{bc} }) (\sqrt{ ( {c}^{2} - 1) ({b}^{2} - 1) } + {2 \sqrt{bc} } )$

c) We want to compute:

$\frac{ \sqrt{9 - 4 \sqrt{5} } }{2 - \sqrt{5} }$

Let the numerator,

$\sqrt{9 - 4 \sqrt{5} } = a - b \sqrt{5}$

Square both sides of the equation;

$9 - 4 \sqrt{5} = {a}^{2} - 2ab \sqrt{5} + 5 {b}^{2}$

Compare coefficients in both equations;

${a}^{2} + 5 {b}^{2} = 9 - - - (1)$

and

$- 2ab = - 4 \\ ab = 2 \\ a = \frac{2}{b} - - - - (2)$

Put equation (2) in (1) and solve;

$\frac{4}{ {b}^{2} } + 5 {b}^{2} = 9$

$5 {b}^{4} - 9 {b}^{2} + 4 = 0$

$b = \pm1$

When b=-1, a=-2

This means that:

$\sqrt{9 - 4 \sqrt{5} } = - 2 + \sqrt{5}$

This implies that:

$\frac{ \sqrt{9 - 4 \sqrt{5} } }{2 - \sqrt{5} } = \frac{ - 2 + \sqrt{5} }{2 - \sqrt{5} } = \frac{ - (2 - \sqrt{5)} }{2 - \sqrt{5} } = - 1$

3 0

2 years ago

Read 2 more answers