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Eduardwww [97]
2 years ago
6

Easy question - giving brainly if correct AND WORK IS SHOWN.​

Mathematics
2 answers:
Sedbober [7]2 years ago
7 0

ANSWER:
The best bargain is C


EXPLANATION:
$12.00 divided by 8 = $1.50
$5.90 divided by 4 = $1.47
$3.10 divided by 2 = $1.55
k0ka [10]2 years ago
5 0

Answer:

hi im new

Step-by-step explanation:

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Help please!!! I just seriously need help I'm bad at math>
azamat
There is only 1 real number solution
8 0
3 years ago
An isosceles triangle has an area of 254 km², and the angle between the two equal sides is 53°. Find the length of the two equal
torisob [31]
Area of a triangle = 1/2 ab sinC

In this equation the angle C is between the two adjacent sides ab
Since the lengths are equal we can call them both X
Hence 254= 1/2 (x)(x) sin53
By simplifying 508 = (x^2) sin53
So therefore x is the square root of 508 divided by sin53 which means the length of the sides is 35.8km to 3 significant figures
8 0
3 years ago
What is this math answer?
qwelly [4]

Answer: me + u is us

Step-by-step explanation:

3 0
3 years ago
What is the area of this figure? A 28 yd² B 40 yd C52 yd² D64 yd²
d1i1m1o1n [39]
<span>D64 yd² i hope this helps</span>
3 0
3 years ago
Read 2 more answers
Lim x-0 (sin2xcsc3xsec2x)/x²cot²4x
sergij07 [2.7K]

By the definitions of cosecant, secant, and cotangent, we have

\dfrac{\sin2x\csc3x\sec2x}{x^2\cot^24x}=\dfrac{\sin2x\sin^24x}{x^2\sin3x\cos2x\cos^24x}

Then we rewrite the fraction as

\dfrac{\sin2x}{2x}\left(\dfrac{\sin4x}{4x}\right)^2\dfrac{3x}{\sin3x}\dfrac{32}{3\cos2x\cos^24x}

The reason for this is that we want to apply the well-known limit,

\displaystyle\lim_{x\to0}\frac{\sin ax}{ax}=\lim_{x\to0}\frac{ax}{\sin ax}=1

for a\neq0. So when we take the limit, we have

\displaystyle\lim_{x\to0}\cdots=\lim_{x\to0}\frac{\sin2x}{2x}\left(\lim_{x\to0}\frac{\sin4x}{4x}\right)^2\lim_{x\to0}\frac{3x}{\sin3x}\lim_{x\to0}\frac{32}3\cos2x\cos^24x}

=1\cdot1^2\cdot1\cdot\dfrac{32}3=\dfrac{32}3

8 0
3 years ago
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