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Juli2301 [7.4K]

2 years ago

8

What is the length of line MN?

2 answers:

Elza [17]2 years ago

8 0

Answer:

that is the solution to the question

madreJ [45]2 years ago

4 0

Answer:

5

Step-by-step explanation:

Both side have to equal

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How much will it cost guys

aalyn [17]

Answer:

635 pounds hope you do good :)

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

<u>Prove that:</u>

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

<u>Proof: </u>

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

<u>Taking</u><u> </u><u>LHS</u>

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

<em>Substituting</em> the value of <em>cos A, cos B and cos C,</em>

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

<em>On combining the fractions,</em>

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

<em>Regrouping the terms,</em>

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

2 years ago

A scuba diver returning to the surface swims to move up quickly, then waits a few meters below the surface to avoid pressure iss

Juliette [100K]

Given:

The function f represents the diver's distance from sea level over time, in seconds.

$f(x)=\begin{cases}3x+1 & \text{ if } -9$

To find:

The correct statement for the domain.

Solution:

Domain is the set of input values for which the function is defined.

Here, x values are the input values and they represent time intervals on which the function is defined

Here, the possible intervals for x are $-9$ .

$Domain=(-9,-2]\cup (-2,1]\cup (1,7)$

$Domain=(-9,7)$

Therefore, the domain is the time interval (−9, 7). Hence the correct option is A.

4 0

2 years ago

5. If y=x(x+2)2, the degree of the polynomial is<br>B. 2<br>A. 1<br>D. 4<br>C. 3

stira [4]

Answer:

2

Step-by-step explanation:

x times x is x^2

7 0

2 years ago

Determine if the following side lengths can create a triangle:<br> 5, 7, 13<br> A. Yes<br> B. No

Paul [167]

Answer:

No

Step-by-step explanation:

In order for three sides to make a triangle, if any two sides are added together, their sum should be greater than the third side:

=><u> 5 + 7 = 12 < 13 (DOES NOT check off the rule)</u>

=> 7 + 13 = 20 > 5 (Checks off the rule)

=> 5 + 13 = 18 > 7 (Checks off the rule)

As shown, even though 2 of the statements check off the rule the third one does not, this means that the sides don't make a triangle.

Hope this helps!

8 0

2 years ago

Read 2 more answers