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2 years ago

Please help me with this

Mathematics

2 answers:

Aleksandr-060686 [28]2 years ago

8 0

Answer:

Part A: Let c = the cost of each candy bar

3 + 8c = 43

8c = 40

Part B: c = $5.00

Step-by-step explanation:

hram777 [196]2 years ago

6 0

I know part b would be $5

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Help me guys thx so much

navik [9.2K]

Answer:

Option D, $10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

Step-by-step explanation:

Step 1: Multiply

$(\sqrt{10x^4} -x\sqrt{5x^2} )*(2\sqrt{15x^4} + \sqrt{3x^3})\\ (\sqrt{10 * x^2 * x^2} -x\sqrt{5 * x^2} ) * (2\sqrt{15 * x^2 * x^2} +\sqrt{3 * x^2 * x})\\(x^2\sqrt{10} -x^2\sqrt{5} )*(2x^2\sqrt{15} +x\sqrt{3x}) \\\\$

$(x^2\sqrt{10}*2x^2\sqrt{15} )+(x^2\sqrt{10}*x\sqrt{3x} ) + (-x^2\sqrt{5} *2x^2\sqrt{15}) + (-x^2\sqrt{5} *x\sqrt{3x}$

$(2x^4\sqrt{150} ) + (x^3\sqrt{30x}) + (-2x^4\sqrt{75}) + (-x^3\sqrt{15x} )$

$2x^4\sqrt{5^2*6} + x^3\sqrt{30x} -2x^4\sqrt{5^2*3} -x^3\sqrt{15x}$

$10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

Answer: Option D, $10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

6 0

3 years ago

Read 2 more answers

57.8 ÷ 81 (63 points and too lazy to solve the questions but help!)

Luba_88 [7]

0.71358024691

If you are rounding then the answer would be 0.71

4 0

2 years ago

Read 2 more answers

A random sample of bolts is taken from inventory, and their lengths are measured. The average length in the sample is 5.3 inches

LekaFEV [45]

Answer:

$L_x=5.3 inches$

Step-by-step explanation:

Average length $\=x =5.3 inches$

Standard deviation $\sigma=0.2 inches$

Sample size $n=50$

Generally The point estimate for the mean length of all bolts in inventory is

$L_x= \=x$

$L_x=5.3 inches$

3 0

3 years ago

How to solve this equation algebraically: Y=2x+1 3x+4y=26

AURORKA [14]

You can substitute what y is into the second equation, so:
3x + 4(2x + 1) = 26
3x + 8x + 4 = 26
11x + 4 = 26
- 4
11x = 22
÷ 11
x = 2

y = (2 × 2) + 1
y = 5 + 1
y = 5

So you get x as 2 and y as 5, I hope this helps!

3 0

3 years ago

A bag contains blue and yellow marbles. Two marbles are drawn without replacement. The probability of selecting a blue marble an

exis [7]

blue marbles
y = yellow marbles
Sum = b+y
The chance of a blue marble being drawn first is:
b / (b+y) = 0.55
The chance of a blue marble being drawn first then a yellow next is:
b / (b+y) * y / (b+y-1) = 0.37
This can be solve easily by using a theorem of Bayes
0.37/0.55 = .67 or 67%

7 0

3 years ago