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stich3 [128]
3 years ago
10

9.25 the average score for Mrs Myles class was 70% of Mrs Jones i class test average score in Mr. Myles class

Mathematics
1 answer:
vodka [1.7K]3 years ago
3 0
80.2 is the correct answer
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The tides are meaured at a forth D. The mean of the low tide values at Locations A,B C, and D,s is -0.54. What is the value of t
MAVERICK [17]
A - Difference in tides for location C is 6.9 feet and high tide for location A is 8.2 feet.
B - Value of low tide in location D is -0.64 feet.

The table showing the difference between high and low tides.
The difference in tides for location C = 7.34 - 0.44 = 6.9

We are given, the mean of the low tides for the 4 locations is -0.54
3 0
3 years ago
Manuel is standing 5.3m north of jackets. Rowyn is standing 2.5m west of the jackets . Dennis is standing 1.9m east of the jacke
kkurt [141]
He would be standing 5.3+2.5=7.8

so hes standing 7.8m away 
4 0
3 years ago
How do i solve this? (by factoring)<br> x(x+3)=x+8
AfilCa [17]

\sf x\left(x+3\right)=x+8\\\\\\\\x^2+3x=x+8\\\\\\\\\sf \boxed{\sf\mathrm{Subtract\:}8\mathrm{\:from\:both\:sides}}\\\\\\\\\sf x^2+3x-8=x+8-8\\\\\\\\\sf \boxed{\sf \mathrm{Simplify}}\\\\\\\\\sf x^2+3x-8=x\\\\\\\\\boxed{\sf \mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}}\\\\\\\\\sf x^2+3x-8-x=x-x\\\\\\\\\boxed{\sf \mathrm{Simplify}}\\\\\\\\\sf x^2+2x-8=0\\\\\\\\\boxed{\sf factor~ x^2+2x-8=0}\\\\\\\\\sf \left(x-2\right)\left(x+4\right)=0

\sf \boxed{\sf \mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0}\\\\\\\\\sf x-2=0\quad \mathrm{or}\quad \:x+4=0\\\\\\\\\boxed{\sf \{x=2\}}\\\\\boxed{\sf \{x=-4 \} }

3 0
3 years ago
A woman wants to build a rectangular garden. She plans to use a side of a shed for one side of the garden. She has 84 yd of fenc
Ivan
Let x denote the length of the side of the garden which is covered fenced by a shed, and \frac{A}{x} be the width of the garden.

The perimeter of a rectangle is given by 2(length + width)
i.e. 2x + \frac{A}{x} = 84
which gives:
A = 84x - 2x^2

For the area to be maximum, the differentiation of A with respect to x must be equal to 0.
i.e. \frac{dA}{dx} =84-4x=0 \\ 4x=84 \\ x=21

Therefore, the maximum area of the garden enclosed is given by
84(21)-2(21)^2=1764-2(441)=1764-882=882 \, yd^2

3 0
3 years ago
Find each unit rate. Round to the nearest hundredth if necessary.<br> 13 apples for 2 pies
Sunny_sXe [5.5K]
6.5 apple per pie, is that what the question is asking?
4 0
4 years ago
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