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3 years ago

HELP ASAP PLEASE!! I don’t know the answer plz help me

Mathematics

1 answer:

Vanyuwa [196]3 years ago

5 0

Answer:56 to 96

Step-by-step explanation:

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Find the product. -8x5y2 · 6x2y

katrin [286]

Answer:

The product of the given expression $-8x^5y^2\times 6x^ 2y$ = 72x^6y^3[/tex]

Step-by-step explanation:

Given expression is $-8x^5y^2\times 6x^2y$

To find the product of the given expression:

The product of the given expression can be written as

$-8x^5 y^2\times 6x 2y=(-8x^5y^2)(6x^2y)$

$=(-8x^5y^2)(6x^2y)$

$=-48(x^5\times x^2)(y^2\times y^1)$ [since $a^m\times a^n=a^{m+n}$ ]

$=-48x^7y^3$

Therefore the product of the given expression $-8x^5y^2\times 6x^2y$ is $-48x^7y^3$

Therefore, $-8x^5y^2\times 6x^2y$ =-48x^7y^3[/tex]

3 0

3 years ago

You have 5 55 reindeer, Bloopin, Rudy, Ezekiel, Prancer, and Balthazar, and you want to have 3 33 fly your sleigh. You always ha

sweet-ann [11.9K]

Answer:

60 different ways.

Step-by-step explanation:

We have been given that you have 5 reindeer, Bloopin, Rudy, Ezekiel, Prancer, and Balthazar, and you want to have 3 fly your sleigh. You always have your reindeer fly in a single-file line. We are asked to find the number of ways in which we can arrange reindeer.

We will use permutations to solve our given problem.

The number of way to choose r objects from a set of n objects is $^nP_r=\frac{n!}{(n-r)!}$ .

Upon substituting our given values in permutations formula, we will get:

$^5P_3=\frac{5!}{(5-3)!}$

$^5P_3=\frac{5!}{2!}$

$^5P_3=\frac{5\times 4\times 3\times 2!}{2!}$

$^5P_3=5\times 4\times 3$

$^5P_3=60$

Therefore, you can arrange your reindeer in 60 different ways.

4 0

3 years ago

How many grams are in 1 pound (on Earth)? Round to the nearest tenth.

cupoosta [38]

Answer:

453.6, or 453.59237

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Please someone help me to prove this..

Pachacha [2.7K]

Answer: see proof below

<u>Step-by-step explanation:</u>

Use the following Sum to Product Identities:

$\sin x+\sin y=2\sin \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\sin x-\sin y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=-2\sin \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)$

<u>Proof LHS → RHS</u>

$\text{LHS:}\qquad \qquad \qquad \dfrac{\sin 5-\sin 15+\sin 25 - \sin 35}{\cos 5-\cos 15- \cos 25 + \cos 35}$

$\text{Reqroup:}\qquad \qquad \qquad \dfrac{(\sin 25+\sin 5)-(\sin 35 + \sin 15)}{(\cos 35+\cos 5)-(\cos 25 + \cos 15)}$

$\text{Sum to Product:}\quad \dfrac{2\sin \bigg(\dfrac{25+5}{2}\bigg)\cos \bigg(\dfrac{25-5}{2}\bigg)-2\sin \bigg(\dfrac{35+15}{2}\bigg)\cos \bigg(\dfrac{35-15}{2}\bigg)}{2\cos \bigg(\dfrac{25+15}{2}\bigg)\cos \bigg(\dfrac{25-15}{2}\bigg)-2\cos \bigg(\dfrac{35+5}{2}\bigg)\cos \bigg(\dfrac{35-5}{2}\bigg)}$ $\text{Simplify:}\qquad \qquad \dfrac{2\sin 15\cos 10-2\sin 25\cos 10}{2\cos 20\cos 15-2\cos 20\cos 5}$

$\text{Factor:}\qquad \qquad \dfrac{2\cos 10(\sin 15-\sin 25)}{2\cos 20(\cos 15-\cos 5)}$

$\text{Sum to Product:}\qquad \dfrac{\cos 10\bigg[2\cos \bigg(\dfrac{15+25}{2}\bigg)\sin \bigg(\dfrac{15-25}{2}\bigg)\bigg]}{\cos 20\bigg[-2\sin \bigg(\dfrac{15+5}{2}\bigg)\sin \bigg(\dfrac{15-5}{2}\bigg)\bigg]}$

$\text{Simplify:}\qquad \qquad \dfrac{\cos 10[2\cos 20\sin (-5)]}{\cos 20[-2\sin 10\sin 5]}\\\\\\.\qquad \qquad \qquad =\dfrac{-2\cos10 \cos 20 \sin 5}{-2\sin 10 \cos 20 \sin 5}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 10}{\sin 10}\\\\\\.\qquad \qquad \qquad =\cot 10$

LHS = RHS: cot 10 = cot 10 $\checkmark$

8 0

3 years ago

PLEASE ANSWET FAST!!!

Mazyrski [523]

Answer:

30 ladoos

Step-by-step explanation:

8 0

3 years ago