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2 years ago

Write an equation in point-slope form of the line having the given slope that contains the given point. m=34m=34, (36,24)

Mathematics

1 answer:

Basile [38]2 years ago

3 0

All we have to do is input the x, y, and m value into the equation. It'd look like tihs:

y - 24 = 34(x -36)

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The data modeled by the box plots represent the battery life of two different brands of batteries that Mary tested. A) What is t

zhannawk [14.2K]

Answer:

Part A:

Median value for Brand X = 13

Median value for Brand Y = 16

Part B: Brand Y has a longer battery life

Step-by-step explanation:

Part A:

Median value is depicted on a box plot by the vertical line that divides the rectangular box. Therefore:

Median value for Brand X = 13

Median value for Brand Y = 16

Part B:

Brand Y has a higher median value (16) than Brand X (13).

This implies that brand Y has a battery life that last longer than brand X.

8 0

1 year ago

Read 2 more answers

I need what is the slope of the line graphed

pentagon [3]

Answer:

-3/5

Step-by-step explanation:

3 0

2 years ago

Each pair of polygons below is similar. Find the value of x. Some answers may be decimals.

sergiy2304 [10]

Answer:

The answer to your question is below.

Step-by-step explanation:

14.- Use proportions to solve these problems.

$\frac{x + 6}{10} = \frac{x + 9}{14}$

14(x + 6) = 10(x + 9)

14x + 84 = 10x + 90

14x - 10x = 90 - 84

4x = 6

x = 6/4

x = 3/2

15.- $\frac{x - 1}{8} = \frac{x + 2}{10}$

10(x - 1) = 8(x + 2)

10x - 10 = 8x + 16

10x - 8x = 16 + 10

2x = 26

x = 26 / 2

x = 13

16.- $\frac{x + 1}{9} = \frac{10/3}{5}$

5(x + 1) = 9(10/3)

5x + 5 = 30

5x = 30 - 5

5x = 25

x = 25 / 5

x = 5

7 0

2 years ago

Read 2 more answers

3x + 3(x + y) ? Help!

tatuchka [14]

Answer:

=6x+3y

Step-by-step explanation:

3X+3(X+Y)

Distribute:

=3X+(3)(X)+(3)(Y)

=3X+3X+3Y

Combine Like Terms:

=3x+3x+3y

=(3x+3x)+(3y)

=6x+3y

4 0

2 years ago

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

IRISSAK [1]

Prove that:

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

Proof: 

We know that, by Law of Cosines,

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

Taking LHS

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

Substituting the value of cos A, cos B and cos C,

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

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

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

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

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

On combining the fractions,

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

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

Regrouping the terms,

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

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

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

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

2 years ago