3 years ago

Find sin A in triangle ABC

Mathematics

1 answer:

arsen [322]3 years ago

4 0

Sin (A) = BC/AB = 21/75= 7/25

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\frac{1}{5}k-9=\frac{3}{5}k-7

Oksi-84 [34.3K]

let start by multiplying both sides by the LCD of all denominators, in this case that'd be 5, to do away with the denominators.

$\bf \cfrac{1}{5}k-9=\cfrac{3}{5}k-7\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{5}}{5\left( \cfrac{1}{5}k-9 \right)=5\left( \cfrac{3}{5}k-7 \right)}\implies k-45=3k-35 \\\\\\ -45=2k-35\implies -10=2k\implies \cfrac{-10}{2}=k\implies -5=k$

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3 years ago

jenn made 36 candy apples for the fall bake sale. she put 3 candy apples in each bag. jenn sold the bags for $8.00 each. how muc

adelina 88 [10]

$96

36/3 = 12

12*8=96

so she would make $96

7 0

3 years ago

-150.1275 rounded to the nearest tenth

Bingel [31]

I believe that it's -150.1

4 0

4 years ago

Read 2 more answers

Which is the correct form of the quadratic formula? A) x = b ± b2 - 4ac a B) x = b ± b2 - 4ac 2a C) x = - b ± b2 - 4ac a D) x =

DochEvi [55]

That's pretty ill-formatted but the answer is D

$ax^2 + bx + c = 0$

has roots given by

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Bonus quadratic formulas:

The Shakespeare Quadratic Formulas ( $2b$ or $-2b$ ) are:

$x^2 - 2bx + c \textrm{ has zeros at }x=b \pm \sqrt{b^2-c}$

$ax^2 - 2bx + c \textrm{ has zeros at }x=\frac 1 a( b \pm \sqrt{b^2-ac})$

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4 years ago

Since we have computer algebra systems that can solve polynomial division problems for us, why is it necessary to learn how to d

PSYCHO15rus [73]

Answer:

Probably because learning to do polynomials by hand boosts understanding of the topic and improves general algebraic solving ability,both of which are required in further topics like advanced calculus and algebra

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3 years ago