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vlabodo [156]
3 years ago
14

What is the answer to 15 - 5x??​

Mathematics
2 answers:
anzhelika [568]3 years ago
6 0

Answer:

x=3

Step-by-step explanation:

barxatty [35]3 years ago
6 0
X=3 if not please back
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Please help me I’m terrible at math
irga5000 [103]

Answer:

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Step-by-step explanation:

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3 years ago
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Step-by-step explanation:

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2 years ago
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Please help!! How do you do number 1?
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Do a ratio tall over shadow. How tall is tree / shadow. How tall is person / shadow then cross multiply. X/7.5 =5/3. 37.5=3x. Divide by 3 answer 12.5
6 0
3 years ago
Guys!! anyone can help me solve this geometry problem?<br>What is the area of this triangle?
NeX [460]

Answer:

10657.5

Step-by-step explanation:

<h2>Long way that is unnecessarily long</h2>

We can start by finding the area of the larger triangle. Using the Pythagorean theorem, we can say that 251²-105²=the bottom side², and  251²-105²=51976, so the bottom side of the larger triangle is √51976 , or approximately 228. Then, the area of the larger triangle is √51976 * 105/2 = 11969 (approximately). Then, the area of the smallest triangle (the largest triangle - the one that we're trying to find the area of) is 105*(√51976-203)/2 = approximately 1312. Then, subtracting that from the total area, we get (√51976 * 105 -  105*(√51976-203))/2 = 105*203/2 = 10657.5

<h2>Short way</h2>

ALTERNATIVELY, upon further review, we can just see that the height is 105 and the base is 203, so we multiply those two and divide by 2, as is the formula for the area of a triangle, to get 10657.5

3 0
3 years ago
Use the Rational Zeros Theorem to write a list of all potential rational zeros f(x) = 14x^3 + 56x^2 + 2x - 7
Crank
The factors of 7are -1 and 7 or 1 and -7, the factors of 14 are 1, 2, 7, and 14, or -1,  -2, -7,-14. so the list of potential zeros are: 1/1, 1/2, 1/7, 1/14, 7/1,7/2, 7/7, 7/14, which can be simplified into 1, 1/2,1/7, 1/14, 7, 7/2
add the negative ones: -1, -1/2,-1/7, -1/14, -7, -7/2
I believe there are a total of 12 potential zeros
reference: 
http://www.sparknotes.com/math/algebra2/polynomials/section4.rhtml

3 0
3 years ago
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