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3 years ago

Help pleaseeeeeeeeeeseeeeeeeeeeww

Mathematics

2 answers:

Sedbober [7]3 years ago

7 0

Answer:

Step-by-step explanation:

The interior angles of triangles always add up to 180. That means that you can just add 64 to 90, which equals 154. Then subtract 154 from 180, which then equals 26.

Hope that this helps!

Tema [17]3 years ago

6 0

Answer:

Step-by-step explanation:

Given Angles=90° and 64°

We know that in a triangle

$\boxed{\sf Sum\:of\:angles=180°}$

$\rm \rightharpoonup \: 90 + 64 + x = 180 \\ \rm \rightharpoonup \: 154 + x = 180 \\ \rm \rightharpoonup \: x = 180 - 154 \\ \rm \rightharpoonup \: x = 26$

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H^2+14h=-31 Solve by completing the square

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Since the constant has been moved to the left side, you can move on to the next step which is adding (b/2)² to both sides of the equation.

h² + 14h + (14/2)² = -31 + (14/2)²

Simplify the parenthesis and exponent.

h² + 14h + 7² = -31 + 7²

h² + 14h + 49 = -31 + 49

h² + 14h + 49 = 18

Factor the expression of the left.

(h + 7)(h + 7) = 18

Take the square root of both sides.

√(h + 7)(h + 7) = ± √9 • 2

(h + 7) = ± 3√2

h + 7 = ± 3√2

Subtract 7 from both sides.

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h = -7 + 3√2 → -2.7573 → -2.76

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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:

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

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

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:

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

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

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Solution set:

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

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