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Svetllana [295]
2 years ago
11

Make x the subject of the formula y= 2x +5

Mathematics
1 answer:
Kryger [21]2 years ago
5 0

Answer:

x= (y-5)/2

Step-by-step explanation:

Isolate the x variable.

y= 2x +5

Subtract 5 from both sides.

y-5=2x

Divide both sides by 2.

(y-5)/2= x

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Write in slope-intercept form an equation of the line that passes through the given points.
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Y=1/3x-3 Is the answer to it
7 0
3 years ago
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Solve the equation without the use of a calculator. please Help!!! Thank you!!!<br> 12x^3-30x=5-2x^2
nata0808 [166]

So, since we have a cubic equation with 4 terms, the first thing we should try is factoring by grouping, so:

12x^3+2x^2-30x+5 =0 \implies \\ 2x^2(6x+1)-5(6x+1)=0 \implies \\ (2x^2-5)(6x+1) = 0

Now that we've factored our equation, we can use ZPP and break it up:

2x^2-5 = 0 \implies \\ 2x^2=5 \implies\\ x^2=\frac{5}{2} \implies\\ x=\pm\frac{\sqrt{5}}{\sqrt{2}}=\pm\frac{\sqrt{10}}{2} \\ \\ 6x+1 =0 \implies \\ 6x=-1 \implies\\ x=\frac{-1}{6}

So, our solutions are:

x\in \{\frac{\sqrt{10}}{2},-\frac{\sqrt{10}}{2},-\frac{1}{6}\}

4 0
3 years ago
!!PLS HELP I WILL GIVE BRAINLEST!!
shutvik [7]

they amount of money are not the same

reason why;

the reason why its not the same is because 2x8x9=144 and 3x9x12=324 and the amount of money spent is different

3 0
2 years ago
Solve the equation by taking square roots. Problem: 8b^2-7=-76
yulyashka [42]

Answer:

b = +5

Step-by-step explanation:

8b^2=193+7

8b^2=200

b^2= 200/8

b^2=25

b=√25

b=+5

6 0
3 years ago
The circular stream of water from a faucet is observed to taper from a diameter of 20 mm to 10 mm in a distance of 50 cm. Determ
steposvetlana [31]

Solution:

From Bernoulli equation

\frac{1}{2}ρV_{1}^{2} + ρgh = \frac{1}{2}ρV_{2}^{2} , where ρ is density of water, h – height difference and V_{1} and V_{2}  are the velocities in upper and lower cross sections correspondingly. We take into account that the pressure in both cross-sections is the same and equal to the atmospheric one.

From the continuity and assuming water incompressible:

\pi r_{1}^{2}V_{1} = \pi r_{2}^{2}V_{2} , where A_{1} and A_{2} are the corresponding cross-sections. As the lower diameter is twice as low as the upper one, we can conclude that V_{1} = \frac{V_{2}}{4}

Inserting it back into Bernoulli equations produces:

V_{1} = \sqrt{\frac{2}{15}gh} = 0.26m/s and the flow rate is

\pi r_{1}^{2}V_{1} = \pi r_{2}^{2}V_{2} = 8.10^{-5} \frac{m^{3} }{s}

6 0
3 years ago
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