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Vanyuwa [196]
2 years ago
7

Determine the slope from the given graph below:

Mathematics
1 answer:
DochEvi [55]2 years ago
3 0

Answer:

i

Step-by-step explanation:

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Evaluate the promblem. 3x+6+5x=86
Mashcka [7]

Answer:

10 or x=10

Step-by-step explanation:

First we can combine like-terms.

8x+6=86

Then subtract 6 from both sides

8x=80

Divide by 8

x=10

8 0
3 years ago
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Please help and explain
Sonja [21]

Find out how much over the 10 years whatever it says and then use that to explain if it's worth is

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4 years ago
Al trazar en el plano cartesiano el angulo a cuyo lado terminal es el punto de ( 3,4). Los valores de las funciones del coseno y
Igoryamba

Answer:

cos(\alpha)=\frac{3}{5}=0.6

cosec(\alpha)=\frac{5}{4}=1.25

Step-by-step explanation:

El cos(α) se define como el cociente entre el cateto adyacente y la hipotenusa.

El valor del cateto adyacente en nuestgro caso es CA = 3.

La hipotenusa se calcual de la siguiente manera:

h=\sqrt{3^2+4^2}=5

Por lo tanto, el cos(α) sera:

cos(\alpha)=\frac{3}{5}=0.6

El cosec(α)=h/CO.

El cateto opuesto CO = 4 y la hipotenusa h = 5

Por lo tanto, el cosec(α) sera:

cosec(\alpha)=\frac{5}{4}=1.25

Espero te haya sido de ayuda!

4 0
3 years ago
The circumference of a clock is 44 inches. Find the
Brut [27]

Answer:

Click on the picture it explains everything my friend! didnt want to write it out on brianly so yeah. Hope this helps!

4 0
3 years ago
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The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of
Lilit [14]

Check the picture below.

\bf (\stackrel{a}{1}~,~\stackrel{b}{-1})\qquad \impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c = \sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{1^2+(-1)^2}\implies c=\sqrt{2} \\\\[-0.35em] ~\dotfill

\bf sin(\theta ) \implies \cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{-1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies -\cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies -\cfrac{\sqrt{2}}{2}

\bf cos(\theta ) \implies \cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies \cfrac{\sqrt{2}}{2} \\\\\\ tan(\theta ) = \cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{1}}\implies tan(\theta ) = -1

7 0
3 years ago
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