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3 years ago

PLSS HELPP 26 POINTS AND BRAINLIEST

Mathematics

1 answer:

belka [17]3 years ago

5 0

Answer:

$\Large \boxed{\sf 150 \ m^2}$

Step-by-step explanation:

Surface area of this shape = area of 2 triangles + area of 3 rectangles

$(12 \times 5 \times 0.5 \times 2 )+(13 \times 3+12 \times 3+5 \times 3)= 150$

Height of the triangular base is 12 m.

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40 min + 1 hour + 35 minutes =.........h.......min......

Ratling [72]

Answer:

2 hour 15 min

Step-by-step explanation:

60 min = 1 hour

1 hour + 35 min + 40 min

1 hour + 75 min

1 hour + 1 hour + 15 min

= 2 hour 15 min

3 0

3 years ago

PLEASE HELP

algol13

Independent events mean that one event happening does not have to have the other event to happen.

The probability of both events happening is the product of the events.

The answer would be:

E. P(A ∩ B) = P(A) x P(B)

8 0

4 years ago

Read 2 more answers

What is 12 X -19 equals 5X -5

Anettt [7]

Answer:

Step-by-step explanation:12x=5x+14, 7x=14,

x=2

5 0

3 years ago

HELP!!! this is graded and no links plzzzzz!!!!!

vampirchik [111]

Answer:

88cm^2

Step-by-step explanation:

Area of the rectangle: base x height

Area of parallelogram: base x height

Area of the shaded part: area of rectangle -area of parallelogram:

12x10-4x8=120-32=88

5 0

3 years ago

Can someone help me out ! got stuck in this question for an hour.

Reil [10]

Answer:

See below

Step-by-step explanation:

Considering $\vec{u}, \vec{v}, \vec{w} \in V^3 \lambda \in \mathbb{R}$ , then

$\Vert \vec{u} \cdot \vec{v}\Vert \leq \Vert\vec{u}\Vert \Vert\vec{v}\Vert$ we have $(\vec{u} \cdot \vec{v})^2 \leq (\vec{u} \cdot \vec{u})(\vec{v} \cdot \vec{v}) \quad$$

This is the Cauchy–Schwarz Inequality, therefore

$\left(\sum_{i=1}^{n} u_i v_i \right)^2 \leq \left(\sum_{i=1}^{n} u_i \right)^2 \left(\sum_{i=1}^{n} v_i \right)^2$

We have the equation

$\dfrac{\sin ^4 x }{a} + \dfrac{\cos^4 x }{b} = \dfrac{1}{a+b}, a,b\in\mathbb{N}$

We can use the Cauchy–Schwarz Inequality because $a$ and $b$ are greater than 0. In fact, $a>0 \wedge b>0 \implies ab>0$ . Using the Cauchy–Schwarz Inequality, we have

$\dfrac{\sin ^4 x }{a} + \dfrac{\cos^4 x }{b} =\dfrac{(\sin^2 x)^2}{a}+\dfrac{(\cos^2 x)}{b}\geq \dfrac{(\sin^2 x+\cos^2 x)^2}{a+b} = \dfrac{1}{a+b}$

and the equation holds for

$\dfrac{\sin^2{x}}{a}=\dfrac{\cos^2{x}}{b}=\dfrac{1}{a+b}$

$\implies\quad \sin^2 x = \dfrac{a}{a+b} \text{ and }\cos^2 x = \dfrac{b}{a+b}$

Therefore, once we can write

$\sin^2 x = \dfrac{a}{a+b} \implies \sin^{4n}x = \dfrac{a^{2n}}{(a+b)^{2n}} \implies\dfrac{\sin^{4n}x }{a^{2n-1}} = \dfrac{a^{2n}}{(a+b)^{2n}\cdot a^{2n-1}}$

It is the same thing for cosine, thus

$\cos^2 x = \dfrac{b}{a+b} \implies \dfrac{\cos^{4n}x }{b^{2n-1}} = \dfrac{b^{2n}}{(a+b)^{2n}\cdot b^{2n-1}}$

Once

$\dfrac{a^{2n}}{(a+b)^{2n}\cdot a^{2n-1}}+ \dfrac{b^{2n}}{(a+b)^{2n}\cdot b^{2n-1}} =\dfrac{a^{2n}}{(a+b)^{2n} \cdot \dfrac{a^{2n}}{a} } + \dfrac{b^{2n}}{(a+b)^{2n}\cdot \dfrac{b^{2n}}{b} }$

$=\dfrac{1}{(a+b)^{2n} \cdot \dfrac{1}{a} } + \dfrac{1}{(a+b)^{2n}\cdot \dfrac{1}{b} } = \dfrac{a}{(a+b)^{2n} } + \dfrac{b}{(a+b)^{2n} } = \dfrac{a+b}{(a+b)^{2n} }$

dividing both numerator and denominator by $(a+b)$ , we get

$\dfrac{a+b}{(a+b)^{2n} } = \dfrac{1}{(a+b)^{2n-1} }$

Therefore, it is proved that

$\dfrac{\sin ^{4n} x }{a^{2n-1}} + \dfrac{\cos^{4n} x }{b^{2n-1}} = \dfrac{1}{(a+b)^{2n-1}}, a,b\in\mathbb{N}$

4 0

3 years ago