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baherus [9]
3 years ago
11

Consider the polynomial function h(x) = -2x^5+ 8x^4

Mathematics
1 answer:
kvv77 [185]3 years ago
3 0

Answer: B

Step-by-step explanation:

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Find the LCM<br> of 9 and 27.
Mnenie [13.5K]

Answer:

27

Step-by-step explanation:

The answer is 27, because 9 x 3 = 27 and the smallest number that is the same is 27.

Hope this helps!

6 0
3 years ago
Read 2 more answers
You have a Bread recipe:- 500 g flour 2 teaspoons salt 7 g dry yeast 4 tablespoons olive oil 300 ml water If you’re scaling down
makvit [3.9K]

Answer:

228 grams

Step-by-step explanation:

500/2=250

7*4=28

300-250=50

250-50-+28=228

6 0
3 years ago
10. Three kinds of teas are worth $4.60 per pound, $5.75 per pound, and $6.50 per pound. They are to be
zepelin [54]

Answer:

The mass of the $4.60/lb tea that should be used in the mixture is 10 lb

The mass of the $5.75/lb tea that should be used in the mixture is 8 lb

The mass of the $6.50/lb tea that should be used in the mixture is 2 lb

Step-by-step explanation:

The parameters of the question are;

The worth of the three teas are

Tea A = $4.60/lb

Tea B = $5.75/lb

Tea C = $6.50/lb

The mass of the mixture of the three teas = 20 lb

The worth of the mixture of the three teas = $5.25 per pound = $5.25/lb

The amount of the $4.60 in the mixture = The sum of the amount of the other two teas

Therefore, given that the mass of the mixture = 20 lb, we have in the mixture;

The mass of tea A + The mass of Tea B + The mass of Tea C = 20 lb

The mass of tea A = The mass of Tea B + The mass of Tea C

Therefore;

The mass of tea A + The mass of tea A = 20 lb

2 × The mass of tea A in the mixture = 20 lb

The mass of tea A in the mixture = 20 lb/2 = 10 lb

The mass of tea A in the mixture = 10 lb

The mass of Tea B + The mass of Tea C = The mass of tea A = 10 lb

The mass of Tea B + The mass of Tea C = 10 lb

The mass of Tea B  = 10 lb - The mass of Tea C

Where the mass of Tea C in the mixture = x, we have;

The mass of Tea B in the mixture = 10 lb - x

The cost of the 10 lb of tea A = 10 × $4.60 = $46.0

The worth of the tea mixture = 20 × $5.25 = $105

The worth of the remaining 10 lb of the mixture comprising of tea A and tea B is given as follows;

The worth of Tea B + The worth of Tea C in the mixture = $105.00 - $46.00 = $59.00

Therefore, we have;

x lb × $6.50/lb + (10 - x) lb × $5.75/lb = $59.00

x × $6.50 - x × $5.75 + $57.50 = $59.00

x × $0.75 = $59.00 -  $57.50 = $1.50

x =  $1.50/$0.75 = 2 lb

∴ The mass of Tea C in the mixture = 2 lb

The mass of Tea B in the mixture = 10 lb - x = 10 lb - 2 lb = 8 lb

The mass of Tea B in the mixture = 8 lb

Therefore, since we have;

Tea A = $4.60/lb

Tea B = $5.75/lb

Tea C = $6.50/lb

The mass of tea A in the mixture = 10 lb

The mass of tea B in the mixture = 8 lb

The mass of tea C in the mixture = 2 lb, we find;

The mass of the $4.60/lb tea that should be used in the mixture = 10 lb

The mass of the $5.75/lb tea that should be used in the mixture = 8 lb

The mass of the $6.50/lb tea that should be used in the mixture = 2 lb.

6 0
3 years ago
Given the relation A = {(5,2), (7,4), (9,10), (x, 5)}. Which of the following values for x will make relation A a function?
bija089 [108]
It will be the first one 9
5 0
3 years ago
(12a^(2) - 3)/(2)*(2a + 1)^(-2)*((6)/((2a + 1)))^(-1)
Elena L [17]

Answer:

\frac{144a^{4}+144a^{3} - 36 - 12a} {2a+1}\\

Step-by-step explanation:

\frac{(12a^2-3)}{2*(2a+1)^{-2}}*\frac{6}{(2a+1)^{-1} }\\ = \frac{(12a^2-3)}{2 * \frac{1}{(2a+1)^{2} }} * \frac{6}{\frac{1}{2a+1}  }\\ =\frac{(12a^2-3)(2a+1)^{2}}{2} * \frac{6}{2a+1}\\=\frac{(12a^2-3)(2a+1)^{2}}{2} * \frac{6}{2a+1}

= \frac{(12a^2-3)(4a^{2} + 1 + 2 (2a)(1))}{2} * \frac{6}{2a+1}\\= \frac{(12a^2-3)(4a^{2} + 1 + 4a)}{2} * \frac{6}{2a+1}\\= \frac{(12a^2-3)(4a^{2} + 1 + 4a)}{1} * \frac{3}{2a+1}\\= \frac{3(12a^2-3)(4a^{2} + 1 + 4a)} {2a+1}\\\\= \frac{3(12a^2(4a^{2} + 1 + 4a) - 3 (4a^{2}+1+4a)} {2a+1}\\= \frac{3(48a^{4}+12a^{2}+48a^{3}  - 12a^{2} -12 - 4a)} {2a+1}\\= \frac{144a^{4}+36a^{2}+144a^{3}  - 36a^{2} - 36 - 12a)} {2a+1}

= \frac{144a^{4}+36a^{2}+144a^{3}  - 36a^{2} - 36 - 12a} {2a+1}\\= \frac{144a^{4}+144a^{3} - 36 - 12a} {2a+1}\\

6 0
3 years ago
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