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Ivan
2 years ago
7

SOMEONE PLEASE HELP IF U GET IT RIGHT ILL GIVE U 15 POINTS PLEASE

Mathematics
1 answer:
Novosadov [1.4K]2 years ago
7 0

Answer:

There is no solution, y-int of one line is positive

Step-by-step explanation:

There is no solution because there are no shared points between both lines, and the y intercept of the blue line is positive, because it is above the x-axis.

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The lengths of the diagonals of a rhombus are 4x and 8x. What algebraic expression gives the perimeter of the rhombus?
Marizza181 [45]

Answer:

Approximately 17.88x or 8x\sqrt{5}

Step-by-step explanation:

Use pythagorean formula. In a rhombus the diagonals bisect each other and they are perpendicular, so you could have a right triangle with legs of 2x and 4x, the hypoteneuse would then be \sqrt{(2x)^{2} +(4x)^{2} }= \sqrt{4x^{2} +16x^{2}} which is approximately 4.47x. In a rhombus all 4 sides are the same, so multiply that by 4 and you get the perimeter. 4(4.47x) = 17.88x  or if you simplify the radical instead it's 8x\sqrt{5}

4 0
3 years ago
Plz help! read the instructions and write a proper answer and if you don't i will report you! And the first one to answer right
ahrayia [7]

Answer:

1 = 0.61, 2 = 1.13

Step-by-step explanation:

The first is 25 + 25 + 10 + 1. This is 61 cents which is 0.61 of 1 dollar. The second is 100 + 10 + 1 + 1 + 1. This is equal to 113, 0r 1.13

BRAINLIEST PLEASE

3 0
2 years ago
State whether each pair of terms are like terms. (so yes or no)
Rom4ik [11]

Answer: 4xy and 3xy are, 2s and 5s are, but -10 and -10b aren't.

Step-by-step explanation: 4xy and 2xy have the same terms. 2s and 5s have the same terms but -10a and -10b have different variables.

7 0
3 years ago
A tank has the shape of an inverted circular cone (point at the bottom) with height 10 feet and radius 4 feet. The tank is full
german

Answer:

Step-by-step explanation:

Given that tank has the shape of inverted circular cone

Height = 10 feet and radius =  4 feet

After water pumped out height = 5 ft

Hence volume of water pumped out = \frac{1}{3} \pi R^2 h-\frac{1}{3} \pi r^2 h'

Here we have r/h is constant always

Hence \frac{r}{h'} =\frac{R}{h} \\r=Rh'/h = 4(5)/10 = 2

Substitute to get volume of water pumped out = \frac{1}{3} \pi (10*16-4*5)=\frac{140\pi}{3}

Mass of water = density x volume = 62.4(\frac{140\pi}{3})\\= 2912\pi lbs

Work done = force x displacement

= mass x accen x displacement

Here acceleration = gravity = 32.2 ft/sec^2

Displaement = height reduced = 5 ft

W=2912(32.2)(5)\pi =468832\pi foot -pound

6 0
3 years ago
Evaluate <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bu%7D%7Bz%7D%20%2Bxy%5E3" id="TexFormula1" title="\frac{u}{z} +xy^3" alt="\f
Elanso [62]
The explanation is on the board just tell me if you have questions

8 0
3 years ago
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