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vivado [14]
2 years ago
11

Determine if the given statement is sometimes, always, or never true:

Mathematics
1 answer:
sineoko [7]2 years ago
6 0
If two parallel lines are cut by a transversal - alternate interior angles are always congruent.

However this doesn’t mean always in this case. So sometimes!
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Please help with number 5?
Flauer [41]
Unsure how your teacher wants the table to be filled out, but the solution is about 1.62.

x = 1.75, f(x) =1.0625, g(x) =.75 \\  \\ 
x = 1.64, f(x) =.6896, g(x) =.64 \\  \\ 
x = 1.63, f(x) =.6569, g(x) =.63 \\  \\ 
x = 1.62, f(x) =.6244, g(x) = .62 \\  \\ 
x = 1.61, f(x) = .5921, g(x) = .61
8 0
3 years ago
Simply the expression.
irina1246 [14]
The answer is ctually 44 
68-2 (3x1)4
68-2x3x4
68-24
44
4 0
3 years ago
Philipe
UkoKoshka [18]
<h3>Philipe has to make a weekly sales of $ 3000 so that he can earn $ 360</h3>

<em><u>Solution:</u></em>

Given that,

Philipe  works for a computer store that pays a 12% commission and no salary

We have to find his weekly sales have to be for him to earn $360

From given, we can say,

12 % of weekly sales will earn him $ 360

12 % of weekly sales = 360

Let "x" be the weekly sales

12 % of "x" = 360

Solve the above expression

\frac{12}{100} \times x = 360\\\\0.12x = 360\\\\x = \frac{360}{0.12}\\\\x = 3000

Thus Philipe has to make a weekly sales of $ 3000 so that he can earn $ 360

6 0
3 years ago
In rectangle KLMN, KM = 10x + 24 and LN = 64. Find the value of x.
Mice21 [21]
_________X=30____
_________-________
5 0
3 years ago
A family wants to build a rectangular garden on one side of a barn. If 600 feet of fencing is available to use, then what is the
vovangra [49]

Answer: (A) A=300l-l^{2}

               (B) Length varies between 1 and 150

               (C) Largest area is 22500ft²

Step-by-step explanation: Suppose length is l and width is w.

The rectangular garden has perimeter of 600ft, which is mathematically represented as

2l+2w=600

Area of a rectangle is calculated as

A=lw

Now, we have a system of equations:

2l+2w=600

A=lw

Isolate w, so we have l:

2w=600-2l

w = 300 - l

Substitute in the area equation:

A = l(300 - l)

A = 300l - l²

(A) <u>Function of area in terms of length is given by </u><u>A = 300l - l²</u>

(B) The practical domain for this function is values between 1 and 150.

(C) For the largest area, we need to determine the largest garden possible. For that, we take first derivative of the function:

A' = 300 - 2l

Find the values of l when A'=0:

300 - 2l = 0

2l = 300

l = 150

Replace l in the equation:

w = 300 - 150

w = 150

Now, calculate the largest area:

A = 150*150

A = 22500

<u>The largest area the fence can enclose is </u><u>22500ft².</u>

4 0
3 years ago
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