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3 years ago

Name The postulate or theorem you can use to prove the triangles congruent

Mathematics

1 answer:

Zielflug [23.3K]3 years ago

8 0

The answer is HL. HL is a theorem stating that if the hypotenuse and one leg are congruent to another pair of hypotenuse and leg then the triangles themselves are congruent. Hope this helps.

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Sierra's ice cream shop made $45 before she had to close early when the power went out. She realized she lost Two-thirds of a do

Troyanec [42]

Answer:

the profit in the case when the shop is closed for 6 hours is $41

Step-by-step explanation:

The computation of the profit in the case when the shop is closed for 6 hours is shown below:

= Expression of the situation × number of hours in which the shop is closed

= 45 - 2 ÷ 3 × 6 hours

= $41

hence, the profit in the case when the shop is closed for 6 hours is $41

Therefore the third option is correct

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3 years ago

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Lyrx [107]

Answer:

She bought 6 erasers.

Step-by-step explanation:

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3 years ago

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Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

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3 years ago

IF U UNDERSTAND THIS PLS ANSWER WITH NO LINK!!!

Nikitich [7]

Answer:

First Column top:

5. 3. 10

Second Column middle:

9. 7. 2

Third Column bottom:

4. 8. 6

Hence, the grid would be,

5. 3. 10

9. 7. 2

4. 8. 6

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3 years ago

What's the answer to this question ?

Korvikt [17]

I think the answer is B

4 0

4 years ago