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givi [52]
3 years ago
6

Does a triangle with these side lengths exist? 28, 41, 13

Mathematics
2 answers:
Reika [66]3 years ago
8 0

Answer:

No

Step-by-step explanation:

it doesnt add up to 180 degrees

pychu [463]3 years ago
6 0

Use the triangle inequality theorem.

Add two sides together and see if it is greater than the length of the 3rd side:

28 + 41 = 69 which is greater than 13

41 + 13 = 54 which is greater than 28

28 + 13 = 41 this is equal to the 3rd side length of 41 and not greater than 41

Because 28 + 13 is not greater than 41 those 3 lengths cannot make a triangle.

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Find cos(2*ABC) 100POINTS
juin [17]

Answer:

-\dfrac{7}{25}

Step-by-step explanation:

<u>Trigonometric Identities</u>

\cos(A \pm B)=\cos A \cos B \mp \sin A \sin B

<u>Trigonometric ratios</u>

\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}

where:

  • \theta is the angle
  • O is the side opposite the angle
  • A is the side adjacent the angle
  • H is the hypotenuse (the side opposite the right angle)

Using the trig ratio formulas for cosine and sine:

  • \cos(\angle ABC)=\dfrac{3}{5}
  • \sin(\angle ABC)=\dfrac{4}{5}

Therefore, using the trig identities and ratios:

\begin{aligned}\implies \cos(2 \cdot \angle ABC) & = \cos(\angle ABC + \angle ABC)\\\\& = \cos (\angle ABC) \cos (\angle ABC) - \sin(\angle ABC) \sin (\angle ABC)\\\\& = \cos^2(\angle ABC)-\sin^2(\angle ABC)\\\\& = \left(\dfrac{3}{5}\right)^2-\left(\dfrac{4}{5}\right)^2\\\\& = \dfrac{3^2}{5^2}-\dfrac{4^2}{5^2}\\\\& = \dfrac{9}{25}-\dfrac{16}{25}\\\\& = \dfrac{9-16}{25}\\\\& = -\dfrac{7}{25} \end{aligned}

7 0
2 years ago
Read 2 more answers
What are the coordinates of its image, J'KL, if this
shtirl [24]

Answer:

J'(-15,10)\\\\K'(15,10)\\\\L'(0,-20)

Step-by-step explanation:

For this exercise you must remember that the original figure (before a transformation) is called "Pre-Image" and the one obtained after the transformation is called "Image".

 Dilation is defined as a transformation in which the Image and the Pre-Image have the same shape, but different sizes.

If a figure is dilated by a scale factor "k" with respect to the origin, the rule is:

(x,y) → (kx,ky)

In this case, the vertices of the triangle JKL (the Pre-Image) are:

J(-3.2)\\\\K(3,2)\\\\L(0,-4)

Knowing that the scale factor is:

k=5

You get that the vertices of the triangle J'K'L' (Image), are:

J'=(5(-3),5(2))=(-15,10)\\\\K'=(5(3),5(2))=(15,10)\\\\L'=(5(0),5(-4))=(0,-20)

7 0
3 years ago
Read 2 more answers
The formula for the distance travelled over time t and at an average speed v is v⋅t. Amit ran for 40 minutes at a speed of about
lawyer [7]

Answer:

3 1/3 km or 3.33 km (to the hundredth)

Step-by-step explanation:

Using the given formula,

estimated distance = v.t  = velocity * time,

velocity = 5 km/h

time = 40 minutes = \frac{40}{60} hours = \frac{2}{3}  hours

Therefore

distance = 5 km/h * \frac{2}{3}  hours = 10/3 km

= 3.33 km approximately

3 0
3 years ago
Read 2 more answers
Which values of c will cause the quadratic equation –x2 3x c = 0 to have no real number solutions? check all that apply.
nika2105 [10]

The given quadratic equation will not have any real solution for c<-9/4.

The given quadratic equation is:

-x^{2} +3x+c=0

<h3>What is a quadratic equation?</h3>

Any equation of the form ax^{2} +bx+c=0 is called a quadratic equation with a≠0.

In order to have no real solution, the discriminant of a quadratic equation will be less than zero.

D < 0

3^{2} -4(-1)(c) < 0

9+4c < 0

c < -\frac{9}{4}

For c < -\frac{9}{4} the given quadratic equation will have no real solutions.

Hence, the given quadratic equation will not have any real solution for c<-9/4.

To get more about quadratic equations visit:

brainly.com/question/1214333

5 0
2 years ago
Help! What are the answers?!! I'll give a brainlist.​
Vlada [557]
1) D
2) A
3) C
4) B
5) A
6) D
4 0
2 years ago
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