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2 years ago

What is bigger 3 yards or 10 feet

Mathematics

2 answers:

Studentka2010 [4]2 years ago

4 0

10 feet is bigger than 3 yards I think not so positive

miskamm [114]2 years ago

4 0

There are 3 feet in a yard.
3 feet times 3 yards is 9

9 feet is less than 10 feet.

10 feet is bigger than 3 yards

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I have to use trigonometric identities to solve. But I’m having trouble finding the values of cos A and sin B. Can anyone help m

katrin [286]

let's notice something, angles α and β are both in the I Quadrant, and on the first quadrant the x-coordinate/cosine and y-coordinate/sine are both positive.

$\bf \textit{Sum and Difference Identities} \\\\ cos(\alpha - \beta)= cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(\alpha)=\cfrac{\stackrel{opposite}{15}}{\stackrel{hypotenuse}{17}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}$

$\bf \pm\sqrt{17^2-15^2}=a\implies \pm\sqrt{64}=a\implies \pm 8 = a\implies \stackrel{I~Quadrant}{\boxed{+8=a}} \\\\[-0.35em] ~\dotfill\\\\ cos(\beta)=\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}$

$\bf \pm\sqrt{5^2-3^2}=b\implies \pm\sqrt{16}=b\implies \pm 4=b\implies \stackrel{\textit{I~Quadrant}}{\boxed{+4=b}} \\\\[-0.35em] ~\dotfill$

$\bf cos(\alpha - \beta)=\stackrel{cos(\alpha)}{\left( \cfrac{8}{17} \right)}\stackrel{cos(\beta)}{\left( \cfrac{3}{5} \right)}+\stackrel{sin(\alpha)}{\left( \cfrac{15}{17} \right)}\stackrel{sin(\beta)}{\left( \cfrac{4}{5} \right)}\implies cos(\alpha - \beta)=\cfrac{24}{85}+\cfrac{60}{85} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill cos(\alpha - \beta)=\cfrac{84}{85}~\hfill$

5 0

3 years ago

X=2π/3

Marina86 [1]

Answer:

$cos{\frac{2\pi}{3}}=-\frac{1}{2}$ .

Reference angle is $\frac{\pi}{3}$

Step-by-step explanation:

Given the value of x. we have to find the correct value of cosx.

$x=\frac{2\pi}{3}$

Now, we have to find the exact value of $cos{\frac{2\pi}{3}}$

$cos{\frac{2\pi}{3}}=cos(\frac{\pi}{2}+\frac{\pi}{6})$

$=-sin(\frac{\pi}{6})$

$=-\frac{1}{2}$

Now, we have to find the reference angle of $x=\frac{2\pi}{3}$ .

Since the angle $x=\frac{2\pi}{3}$ lies in second quadrant, the reference angle formula is

Reference angle= $\pi-given angle$ .

= $\pi-\frac{2\pi}{3}=\frac{\pi}{3}$

3 0

2 years ago

Y= 6x + 4 and y= 4x - 2

LenKa [72]

Easy
sub since boht equal y, both are equal
6x+4=y=4x-2
6x+4=4x-2
minus 4x
2x+4=-2
minus 4
2x=-6
divide 2
x=-3
sub back
y=4x-2
y=4(-3)-2
y=-12-2
y=-14

solution in (x,y) form is
(-3,-14)

8 0

2 years ago

Read 2 more answers

Explain what it means for a solution of an equation to be extraneous.

HACTEHA [7]

Answer:

Extraneous Solutions An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation. Example 1: Solve for x, 1 x − 2 + 1 x + 2 = 4 (x − 2) (x + 2).

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2 years ago

What set of transformations could be applied to rectangle ABCD to create A″B″C″D″? 'Rectangle formed by ordered pairs A at negat

tigry1 [53]

Answer:

wag kana mag aral lol

Step-by-step explanation:

sungin mo module mo

3 0

2 years ago