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Snowcat [4.5K]
3 years ago
15

326 ÷ 53 Please help me

Mathematics
1 answer:
nataly862011 [7]3 years ago
8 0
The answer to this question is 6.150943396226415
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A math book at the victor valley college bookstore originally cost $84 to purchase if they sell it for $126 what is the markup r
Sergio039 [100]
50% because you first subtract the original price from the new price so 126-84 equals 42 then you divide that by the original price 84 which will give you .5 which turns to 50%
8 0
3 years ago
4(PX+1)=64 what is the value of x in terms of p and what is the value of x when p is -5 ?
Oksana_A [137]

Answer:

x = -3

Step-by-step explanation:

4(PX+1)=64\\\\p= -5

Substitute -5 for p in the given equation and solve

4(-5(x)+1)=64\\\\\mathrm{Divide\:both\:sides\:by\:}4\\\frac{4\left(-5x+1\right)}{4}=\frac{64}{4}\\\\Simplify\\-5x+1=16\\\\\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}\\-5x+1-1=16-1\\\\Simplify\\-5x =15\\\\\mathrm{Divide\:both\:sides\:by\:}-5\\\frac{-5x}{-5}=\frac{15}{-5}\\\\Simplify\\x = -3

4 0
2 years ago
If a rectangular area is rotated in a uniform electric field from the position where the maximum electric flux goes through it t
Greeley [361]
With a uniform electric field, flux go along parallel paths, then flux is therefore proportional to the cosine of the angle rotated.
0 degree rotation => cos(0)=1 => 100% of flux.
60 degrees rotation => cos(60) => 0.5  => how many % of flux?
7 0
2 years ago
Which is the slope of the line y=−3x+2? −3 −2 2 3
Gnesinka [82]

Answer:

-3

Step-by-step explanation:

y=mx+b where m=slope and b=y-intercept.

3 0
3 years ago
Find the expansion of cos x about the point x=0
ycow [4]

Answer:

Cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{1!} + ...

Step-by-step explanation:

We use Taylor series expansion to answer this question.

We have to find the expansion of cos x at x = 0

f(x) = cos x, f'(x) = -sin x, f''(x) = -cos x, f'''(x) = sin x, f''''(x) = cos x

Now we evaluate them at x = 0.

f(0) = 1, f'(0) = 0, f''(0) = -1, f'''(0) = 0, f''''(0) = 1

Now, by Taylor series expansion we have

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + \frac{f''''(a)(x-a)^4}{4!} + ...

Putting a = 0 and all the values from above in the expansion, we get,

Cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{1!} + ...

8 0
3 years ago
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