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2 years ago

10

in a hostel 30 students have enough food for 40 days how many students should leave the hostel so that the food is enough for 10

0 days

1 answer:

Sonbull [250]2 years ago

8 0

Answer:

18 students

Step-by-step explanation:

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Find the value of c that completes the square for x2−15x+c

meriva

Answer:

The value of <em>c</em> is $\frac{225}{4}$ .

Step-by-step explanation:

The perfect square of the difference between two numbers is:

$(x-y)^{2}=x^{2}-2xy+y^{2}$

The expression provided is:

$x^{2}-15x+c$

The expression is a perfect square of the difference between two numbers.

One of the number is <em>x</em> and the other is √<em>c</em>.

Use the above relation to compute the value of <em>c</em> as follows:

$x^{2}-15x+c=(x-\sqrt{c})^{2}\\\\x^{2}-15x+c=x^{2}-2\cdot x\cdot\sqrt{c}+c\\\\15x=2\cdot x\cdot\sqrt{c}\\\\15=2\cdot\sqrt{c}\\\\\sqrt{c}=\frac{15}{2}\\\\c=[\frac{15}{2}]^{2}\\\\c=\frac{225}{4}$

Thus, the value of <em>c</em> is $\frac{225}{4}$ .

5 0

3 years ago

Can someone help me find the equivalent expressions to the picture below? I’m having trouble

miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is $\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$ .

Now we will solve this expression with the help of law of exponents.

$\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}$

$=2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}$

$=2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}$

$=2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}$

$=2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }$ [Option 2]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$ [Option 1]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$

$=(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }$

$=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }$ [Option 3]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }$

$=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}$ [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

6 0

2 years ago

½x + 1 ⅔ subtracted from 1 ½x - 1 ⅓<br> simplify IF NEEDED

Volgvan

Answer:

$( \frac{1}{2} x + 1 \frac{2}{3} ) - (1 \frac{1}{2} x - 1 \frac{1}{3} ) \\ = ( \frac{1}{2} - \frac{3}{2} )x + ( \frac{5}{3} + \frac{4}{3} ) \\ = \frac{ - 2}{2} x + \frac{9}{3} \\ = - x + 3 \\ = 3 - x$

4 0

3 years ago

5x-15y=45 solve for y

shepuryov [24]

Answer:

y=(0,-3)

Step-by-step explanation:

To find the y-intercept, substitute in 0 for x and solve for y

7 0

2 years ago

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How far is 10% of a 2,000 kilometer trip

Dmitry [639]

Answer:200

Step-by-step explanation:

2000x0.1=200

200

6 0

3 years ago

Read 2 more answers