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harkovskaia [24]

3 years ago

6

A triangle is to be dilated with a scale factor of 3.6. If a side of the original triangle is 8, what is the measure of a side o

f the new triangle?

1 answer:

BlackZzzverrR [31]3 years ago

3 0

What you want to do in dilation is multiply the side(s) by the scal factor to get the measurements of the new shape. in this case multiply 8 by 3.6 to get the new side length of 28.8

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In a recent semester at a local university, 490 students enrolled in both General Chemistry and Calculus I. Of these students, 6

posledela

I think its 149/ 490. Not sure, but pretty confident. Hope this helps.

8 0

3 years ago

Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro

Ugo [173]

Answer:

$Side\ B = 6.0$

$\alpha = 56.3$

$\theta = 93.7$

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with $\alpha, \beta, \theta$

[See Attachment for Triangle]

$A = 10cm$

$C = 12cm$

$\beta = 30$

What the question is to calculate the third length (Side B) and the other 2 angles ( $\alpha\ and\ \theta$ )

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

$B^2 = A^2 + C^2 - 2ABCos\beta$

Substitute: $A = 10$ , $C =12$ ; $\beta = 30$

$B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30$

$B^2 = 100 + 144 - 240*0.86602540378$

$B^2 = 100 + 144 - 207.846096907$

$B^2 = 36.153903093$

Take Square root of both sides

$\sqrt{B^2} = \sqrt{36.153903093}$

$B = \sqrt{36.153903093}$

$B = 6.0128115797$

$B = 6.0$ <em>(Approximated)</em>

Calculating Angle $\alpha$

$A^2 = B^2 + C^2 - 2BCCos\alpha$

Substitute: $A = 10$ , $C =12$ ; $B = 6$

$10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 180 - 144 *Cos\alpha$

Subtract 180 from both sides

$100 - 180 = 180 - 180 - 144 *Cos\alpha$

$-80 = - 144 *Cos\alpha$

Divide both sides by -144

$\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}$

$\frac{-80}{-144} = Cos\alpha$

$0.5555556 = Cos\alpha$

Take arccos of both sides

$Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)$

$Cos^{-1}(0.5555556) = \alpha$

$56.25098078 = \alpha$

$\alpha = 56.3$ <em>(Approximated)</em>

Calculating $\theta$

Sum of angles in a triangle = 180

Hence;

$\alpha + \beta + \theta = 180$

$30 + 56.3 + \theta = 180$

$86.3 + \theta = 180$

Make $\theta$ the subject of formula

$\theta = 180 - 86.3$

$\theta = 93.7$

5 0

3 years ago

Help.....................

eduard

Answer:

Answers in Explanation

Step-by-step explanation:

First Question:

$\sqrt{100}$ + [18 ÷ 3 x 4 - 15] - (60 - 7^2 - 1)

$\sqrt{100}$ + [24 - 15] - (60 - 49 - 1)

$\sqrt{100}$ + 9 - 10

10 + 9 - 10

<u>Answer = 9</u>

Second Question:

5x + 2x = 7x

5x^2 + 3x^2 = 8x^2

2x + 3x - x = 4x

2x + 3y + x + y = 3x + 4y

9x - 6x = 3x

-7y + 3x + 4x + 3y = 7x - 4y

-7x^2 + 2x^2 + 9x^2 = 8x^2

(3x^2 + 5x + 4) - (-1 + x^2) = 2x^2 + 5x + 5

(3 + 2x - x^2) + (x^2 + 8x + 5) = 10x + 8

(3x - 4) - (5x + 2) = -2x - 6

(2x^2 + 5x + 3) - (x^2 - 2x + 3) = x^2 + 7x

(3x^2 + 2x - 5) - (2x^2 - x - 4) = x^2 + 3x - 1

Third Question:

17x + 2y

(5x + 12y) + (3x + y) = 8x + 13y

17x - 8x = 9x

2y - 13y = -11y

Answer: 9x - 11y

7 0

3 years ago

Where is the point at (2.5, -2.5) located

Vanyuwa [196]

Step-by-step explanation:

Go 2.5 units to the right from the origin (0,0) for your x coordinate

Go 2.5 units down from the origin (0,0) for your y coordinate

Where the lines meet is your point.

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3 years ago

Solve for X. 96 93 6 3

Temka [501]

93 is the right answer please mark me as brainlest

Step-by-step explanation:

I say the answer support. me mark me as brainlest

4 0

2 years ago