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Gnesinka [82]
4 years ago
10

Simplify negative 8 over 3 divided by negative 2 over 6. . (5 point

Mathematics
1 answer:
grin007 [14]4 years ago
3 0

(-8/3)/(-2/6)

negative times negative equals positive

simplify -2/6 to -1/3 AND FLIP THE -1/3 TO -3/1

-8/3*-3/1

cancel out the 3s and make them 1s

cancel out the negatives and make them positive

8/1 = 8

<h2><u><em>8</em></u></h2>

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Someone help me with maths questions pls about pythagoras, ill give u brainliest
Marat540 [252]

Answer:

11.183 cm and 22.366 cm

Step-by-step explanation:

Lets call the 25 cm hypotenuse C and the other sides A and B.

B=2A

The Pythagorean theorem states that C^2=A^2+B^2 so:

25^2=A^2+B^2

substitute for B

25^2=A^2+(2A)^2

625=A^2+4A^2

625=5A^2

125=A^2

A=11.183 cm

B=22.366 cm

6 0
3 years ago
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
4 years ago
In the accompanying diagram of parallelogram ABCD, LaTeX: m\angle A=x+17 m ∠ A = x + 17 and LaTeX: m\angle C=2x-4 m ∠ C = 2 x −
jek_recluse [69]
Answer:
x = 21

Explanation:
In a parallelogram, each two opposite angles are equal.

In the given parallelogram ABCD, we can note that angles A and C are opposite angles. This means that they have equal measures.

Therefore:
∠A = ∠C
x + 17 = 2x - 4
17 + 4 = 2x - x
x = 21

We can check:
angle A = 21 + 17 = 38°
angle C = 2(21) - 4 = 38°
They are equal

Hope this helps :)

3 0
3 years ago
(3x-8)° (4x-23)° fine the x
Veronika [31]

You cannot find x.

Because both phrases are to the power of 0, we know that it is essentially 1*1

So:

(3x - 8)^0 (4x - 23)^0 = 1

7 0
3 years ago
9.80x10^-3 + 1.60x10^-4 scientific notation
e-lub [12.9K]
<span>0.00996 would be it I just took a quiz like that but good luck</span>
6 0
3 years ago
Read 2 more answers
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