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lesya692 [45]
3 years ago
9

Sandi had the following problem on her math test:6•4+2÷2-7

Mathematics
1 answer:
Yuki888 [10]3 years ago
3 0
Answer: 18

Explanation:
24+1-7
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Round 0.00076511 to one significant figure
FrozenT [24]

Answer: 0.0008

Step-by-step explanation: Hope this helps :)

6 0
3 years ago
Find the centroid of the region bounded by the given curves y=8sin(4x), y=8cos(4x)
Luba_88 [7]
A cosine is just a sine shifted to the left by π/2. A cosine of 4x is shifted to the left by only π/8 because of the factor 4. Sketch them.

The region we're looking for is this sausage-shaped part between the cos and the sin.

The x intercepts are at π/8 for the cosine and π/4 for the sine. The midpoint between them is at (π/8 + π/4)/2 = 3/16π.

The region is point symmetric around the x axis, so the y coordinate of the centroid is 0.

So the centroid is at (3/16π, 0)
5 0
3 years ago
Thoko deposited R300 every month into savings account from january 31,2020 until January 2022.How many.How many R300's did Thoko
Semenov [28]

Answer:

25

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include final January = 1 month

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8 0
2 years ago
The following is the graph of 8y = 2x − 10
Gelneren [198K]
<span>Simplifying 2x + 8y = 10 Solving 2x + 8y = 10
  Solving for variable 'x'. Move all terms containing x to the left,
 all other terms to the right.
  Add '-8y' to each side of the equation. 2x + 8y + -8y = 10 + -8y Combine like terms: 8y + -8y = 0 2x + 0 = 10 + -8y 2x = 10 + -8y Divide each side by '2'. x = 5 + -4y Simplifying x = 5 + -4y</span>
3 0
4 years ago
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What is the solution of the following trig equation: tan(theta)(csc(theta)+2)=0
gtnhenbr [62]
D:\theta\neq\frac{k\pi}{2};\ k\in\mathbb{Z}\\\\tan\theta(csc\theta+2)=0\iff tan\theta=0\ \vee\ csc\theta+2=0\\\\\\tan\theta=0\iff\theta=k\pi\notin D\\\\csc\theta+2=0\\\\csc\theta=-2\\\\\frac{1}{sin\theta}=-2\\\\-2sin\theta=1\ /:(-2)\\\\sin\theta=-\frac{1}{2}

(\theta=-\frac{\pi}{6}+2k\pi\ \vee\ \theta=\frac{7\pi}{6}+2k\pi)-answer
7 0
3 years ago
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