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stiv31 [10]
4 years ago
8

Solve for x. a. x/2 - x + 2/17 = 3

Mathematics
1 answer:
Alex777 [14]4 years ago
6 0

Answer:

x=-98/17

Step-by-step explanation:

1) simplify

(1/2x + -x) + (2/17) = 3 -- combined like terms

2) subtract 2/17 from both sides

-1/2x = 49/17

3) multiply both sides by 2/-1

x = -98/17

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Plz help :)))))))))))
dusya [7]

Given:

b1 = 17 ft

b2 = 20 ft

h = 13 ft

r = 17/2 = 8.5 ft


Area of the Semi-circle = 1/2 π r²

= 1/2 (22/7) (8.5²)

= (11/7) (72.25)

= 113.535 ft²


Area of the trapezium = (h) (b1 + b2) / 2

= (13)(17 + 20)/2

= 240.5 ft²


Total area of the amphitheater stage = Area of Semi-circle + Area of the trazezium = 113.535 + 240.5 = 354.035 ft²

5 0
3 years ago
Please help me with this
choli [55]

Answer:

Hey mate, just subsitute all the G's for -7 and all the H's for 3 in the question. Then put the question number in the blank that the answer matches.

Step-by-step explanation:

3 0
3 years ago
What is log b^b^6x equivalent to (The b is under log and ^b)? How do you know?
Arturiano [62]

Answer: 6x

Work Shown:

For each step, the logs are all base b. This is to save time and hassle of writing tricky notation of having to write the smaller subscript 'b' multiple times. The first rule to use is that log(x^y) = y*log(x) for any base of a logarithm. The second rule is that \log_b(b) = 1 meaning that the log base of itself is 1

log(b^(6x)) = 6x*log(b) .... pull down exponent using the first rule above

log(b^(6x)) = 6x*1 .... use the second rule mentioned

log(b^(6x)) = 6x

4 0
3 years ago
Rafael bought 6 bags of sugar for his restaurant. Each bag weighed 7.4 kilograms. How many kilograms of sugar did he buy ?
Gnom [1K]

Answer:

44.4 is the answer for this problem be sure to mark brainliest if helpful.


8 0
3 years ago
Read 2 more answers
Consider rolling a six-sided die. Let A be the set of outcomes where the roll is an even number. Let B be the set of outcomes wh
topjm [15]

Answer:

See below

Step-by-step explanation:

A = \{2, 4, 6\}

A^c =\{1, 3, 5\}\\B = \{4, 5, 6\}\\B^c = \{ 1,2, 3\}

To prove 1 we have

(A  \bigcup B) = \{2, 4, 6\} \bigcup \{4, 5, 6\} = \{2, 4, 5, 6\}\\(A  \bigcup B) ^c =\bold{\{1,3\}}\\A^c \bigcap B^c = {\{1, 3, 5\}  \bigcap \{1, 2, 3\} =\bold{\{1,3\}}\\

Hence proved

To prove 2

(A \bigcap B) = \{2, 4, 6\} \bigcap \{4, 5, 6\} = \{4, 6\}\\\\(A \bigcap B)^c = \bold{\{1,2, 3, 5\}}\\A^c \bigcup B^c = \{1, 3, 5\} \bigcup  \{1, 2, 3\} = \bold{\{1,2,3,5\}}\

Hence proved

<u>Notes</u>

The complement of a set is the set of all elements <u>not</u> in the set

Here the set of all outcomes is {1, 2, 3,4,5, 6}

So if A = {2, 4, 6} then the complement of A s=is the set of outcomes not in A ie the set {1, 3, 5} which is the set of odd numbers on a die throw

The union of two sets is the set of all elements in both sets without duplication

The intersection of two sets is the set of all elements common to both sets

7 0
2 years ago
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