A bag of chips costs $3.79 including tax Mr.Connor wants to purchase chips for his class and has a $15 budget. Write and inequality to solve for the number of bags of chips Mr.Connor can purchase with his budget A. $15x > $3.79 B.$15x < $3.79 C $3.79 > $15 D. $3.79 < $15 (There is a flat line under all the greater than and less than signs)

Answers

Answer 1

Answer: The inequality equation that could best express the statement will be letter d. 3.79x < 15. This is because there is only 15 budget from Mr. Connor, and his spending must not exceed the budget. Then the price per pack is 3.79, so it is express as 3.79x.

Answer 2

Answer:

Answer:

D. $3.79x\leq 15$

Step-by-step explanation:

We are given that

Cost of one bag of chips including tax=$3.79

Mr.Connor has budget=$15

We have to write an equality to solve for the number of bags of chips can purchased by Mr.Connor with his budget.

Let x be the number of bags of chips

According to question

$3.79x\leq 15$

Hence, option D is true.

Answer:D. $3.79x\leq 15$

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Which structures secrete chemicals utilized for the completion of digestion within the small intestine?

Answers

Answer:

In the small intestine more digestive enzymes act on the food. Pancreatic juice, from the pancreas, and bile, produced in the liver and stored in the gall bladder, continue to break down various parts of the food. They complete the digestion of starches, sugars, and fats. I hope tis helps

Step-by-step explanation:

Solve the equations: (7x+4)/5 + (4x-2)/2=27, (8x+7)/6 + (9x-2)/2=6, 2/(x-4) + 8/(2x-8)=13 and 2/(x+5) - 4/(7x+35).

Answers

We need to solve for the x in the equations given.
(7x+4)/5 + (4x-2)/2=27
[(7x+4)*2 + (4x-2)*5]/ 10 = 27
(14x+8+20x-10)/10 = 27
(14x+8+20x-10) = 270
34x = 272
x = 8
---------------------

(8x+7)/6 + (9x-2)/2=6
[(8x+7) + (9x-2)*3] / 6 = 6
(8x + 7 + 27x - 6) / 6 = 6
(8x + 7 + 27x - 6) = 36
35x = 35
x = 1
----------------------------

2/(x-4) + 8/(2x-8)=13
[(2*2) + 8] / (2x-4) = 13
12 / (2x-4) = 13
13(2x-4) = 12
26x - 52 = 12
26x = 64
x = 64/26
x = 32 / 13
-----------------------------------

2/(x+5) - 4/(7x+35)
[(2*7) - 4] / (7x+35)
(14-4)/(7x+35)
10 / (7x+35)

An alloy weighing 40 lbs. is 10% tin. The alloy was made by mixing a 15% tin alloy and a 8% tin alloy. How many pounds of each alloy (to the nearest tenth) were used to make the 10% alloy?lbs. of the 15% alloy and lbs. of the 8% alloy.

Answers

Let the weight of 15% tin be x.

So the weight of the 8% tin will be: (40 - x), since the total weight is 40 lbs

Using the mixture formula:

w1*p1 + w2*p2 = w*p

Where w1 = weight of the first percentage, p1 = First Percentage

w2 = weight of the second percentage, p2 = Second Percentage

w = Final weight, p1 = Final Percentage

0.15*x + 0.08(40 - x) = 40*0.1

0.15x + 3.2 - 0.08x = 4

0.15x - 0.08x = 4 - 3.2

0.07x = 0.8

x = 0.8/0.07

x ≈ 11.429

Recall the 15% was made with x, which is ≈ 11.429

Recall the 8% was made with (40-x) ≈ (40 - 11.429) ≈ 28.571

So the 10% tin of the 40 lbs was made with ≈11.4 lbs of the 15% tin and ≈28.6 lbs of the 8% tin

x=15% alloy
y=8% alloy
x+y=40
y=40-x
0.15x+0.08y=40*0.10
0.15x+0.08(40-x)=40*0.1
times 100 both sides for easy
15x+8(40-x)=400
15x+320-8x=400
7x+320=400
minus 320 from both sides
7x=80
divide both sides by 7
x=80/7
x+y=40
80/7+y=280/7
y=200/7
aprox
x=11.4lb
y=28.6lb

11.4lb of 15%
28.6lb of the 8%
what????

3sin(2x)-1=0 to solve

Answers

Answer is in the attachment below.

$3\sin (2x)-1=0 \n \n 3\sin(2x)=1 \qquad /:3 \n \n \sin (2x) = (1)/(3) \n \n 2x=\arcsin (1)/(3) +2\pi k\qquad \hbox{or} \qquad 2x= \pi - \arcsin (1)/(3)+2\pi k \qquad /:2 \n \n x_1=(1)/(2)\arcsin (1)/(3)+\pi k \n \n x_2=(\pi)/(2)-(1)/(2)\arcsin (1)/(3)+\pi k$

k is an integer

arcsin (1/3) is approximately 0.34 radians (19.47 degrees)

3(2y-4) - 2 ) = 0.4 (1+y)

Answers

6y-12-2=0.4+0.4y
6y-10-0.4=0.4-0.4+0.4y
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(-9.6=-5.6y)/5.6
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Please Answer ASAP
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