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kodGreya [7K]
3 years ago
5

PLEASE HELP ASAP!!!!!!!!!! EASY MATH 25 PTS

Mathematics
2 answers:
svetoff [14.1K]3 years ago
8 0

The equation has one solution



Harman [31]3 years ago
4 0

its not worth 25 points but ok here u go let's solve your equation step-by-step.

3y−1=13−4y

Step 1: Simplify both sides of the equation.

3y−1=13−4y

3y+−1=13+−4y

3y−1=−4y+13

Step 2: Add 4y to both sides.

3y−1+4y=−4y+13+4y

7y−1=13

Step 3: Add 1 to both sides.

7y−1+1=13+1

7y=14

Step 4: Divide both sides by 7.

7y

7

=

14

7

y=2

Answer:

y=2 so there your answer is this equation has one solution  your welcome

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BlackZzzverrR [31]

Answer:

The correct answer is option b)

50a+10b<1000; 200a+360b < 7200; a> 0; b>0

Step-by-step explanation:

We are given that Two kind of crated cargo namely A and B to be shipped by truck.

<u>Cargo A:</u>

Volume of each crate of cargo A = 50 cubic ft

Weight of each crate of cargo A = 200 pounds

Let number of crates of cargo A to be shipped = a

Total volume of 'a' crates of cargo A = 50a cubic ft

Total weight of 'a' crates of cargo A = 200a pounds

<u>Cargo B:</u>

Volume of each crate of cargo B = 10 cubic ft

Weight of each crate of cargo B = 360 pounds

Let number of crates of cargo B to be shipped = b

Total volume of 'b' crates of cargo B = 10b cubic ft

Total weight of 'b' crates of cargo B = 360b

Total volume allowed in the truck is 1000 cubic ft

Total volume of 'a' crates of Cargo A and Total volume of 'b' crates of Cargo B = 50a+10b cubic ft (This sum should be less than volume of truck so that it can fit in the truck)

So, the inequality becomes

50a+10b ....... (1)

Total weight allowed (load limit) in the truck is 7200 pounds

Total weight of 'a' crates of Cargo A and Total weight of 'b' crates of Cargo B = 200a+360b cubic ft (This sum should be less than volume of truck so that it can fit in the truck)

So, the inequality becomes

200a+360b ....... (1)

And number of crates of cargo A and B are always a positive number.

So, a > 0 and b > 0.

So, the correct answer is option b.

<em>b. 50a+10b<1000; 200a+360b < 7200; a> 0; b>0</em>

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Step-by-step explanation:

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faust18 [17]

<u>Answer-</u>

<em>The height of the prism is</em><em> 6 units</em>

<u>Solution-</u>

As the base of the prism is a hexagon consisting of 2 congruent isosceles trapezoids.

So,

V_{Prism}=Area_{Base}\times Height

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Area_{Base}=2\times \text{Area of the trapezoid}

Also,

\text{Area of the trapezoid}=\dfrac{1}{2}\times \text{Height}\times (\text{Sum of two parallel lines})

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