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3 years ago

Which expression is equal to 728‾‾‾√−563‾‾‾√ ?

Mathematics

2 answers:

svet-max [94.6K]3 years ago

7 0

The answer is The 3rd one

marusya05 [52]3 years ago

7 0

Answer:

The correct option is 3.

Step-by-step explanation:

The given expression is

$7\sqrt{28}-5\sqrt{63}$

It can be written as

$7\sqrt{4\times 7}-5\sqrt{9\times 7}$

$7\sqrt{2^2\times 7}-5\sqrt{3^2\times 7}$

Using product property of radical expressions, we get

$7(\sqrt{2^2}\times \sqrt{7})-5(\sqrt{3^2}\times \sqrt{7})$ $[\because \sqrt{ab}=\sqrt{a}\sqrt{b}]$

$7(2\times \sqrt{7})-5(3\times \sqrt{7})$

$14\sqrt{7}-15\sqrt{7}$

Taking out common factors.

$\sqrt{7}(14-15)$

$\sqrt{7}(-1)$

$-\sqrt{7}$

Therefore the correct option is 3.

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Define systole and diastole

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Systole and Diastole are 2 phases of the cardiac cycle. Systole is when the heart contracts to pump blood out. Diastole is when the heart relaxes after contraction and refills with blood.

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3 years ago

Determine the equation of the line that passes through the given points. (If you have a graphing calculator, you can use the tab

ICE Princess25 [194]

Answer:

b y= 0.6x + 6

Step-by-step explanation:

When put on a calculator, you can see it passes through the points.

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3 years ago

Match the parabolas represented by the equations with their foci.

Elenna [48]

Function 1 $f(x)=- x^{2} +4x+8$

First step: Finding when $f(x)$ is minimum/maximum
The function has a negative value $x^{2}$ hence the $f(x)$ has a maximum value which happens when $x=- \frac{b}{2a}=- \frac{4}{(2)(1)}=2$ . The foci of this parabola lies on $x=2$ .

Second step: Find the value of y-coordinate by substituting $x=2$ into $f(x)$ which give $y=- (2)^{2} +4(2)+8=12$

Third step: Find the distance of the foci from the y-coordinate
$y=- x^{2} +4x+8$ - Multiply all term by -1 to get a positive $x^{2}$
$-y= x^{2} -4x-8$ - then manipulate the constant of y to get a multiply of 4
$4(- \frac{1}{4})y= x^{2} -4x-8$
So the distance of focus is 0.25 to the south of y-coordinates of the maximum, which is $12- \frac{1}{4}=11.75$

Hence the coordinate of the foci is (2, 11.75)

Function 2: $f(x)= 2x^{2}+16x+18$

The function has a positive $x^{2}$ so it has a minimum

First step - $x=- \frac{b}{2a}=- \frac{16}{(2)(2)}=-4$
Second step - $y=2(-4)^{2}+16(-4)+18=-14$
Third step - Manipulating $f(x)$ to leave $x^{2}$ with constant of 1
$y=2 x^{2} +16x+18$ - Divide all terms by 2
$\frac{1}{2}y= x^{2} +8x+9$ - Manipulate the constant of y to get a multiply of 4
$4( \frac{1}{8}y= x^{2} +8x+9$

So the distance of focus from y-coordinate is $\frac{1}{8}$ to the north of $y=-14$
Hence the coordinate of foci is (-4, -14+0.125) = (-4, -13.875)

Function 3: $f(x)=-2 x^{2} +5x+14$

First step: the function's maximum value happens when $x=- \frac{b}{2a}=- \frac{5}{(-2)(2)}= \frac{5}{4}=1.25$
Second step: $y=-2(1.25)^{2}+5(1.25)+14=17.125$
Third step: Manipulating $f(x)$
$y=-2 x^{2} +5x+14$ - Divide all terms by -2
$-2y= x^{2} -2.5x-7$ - Manipulate coefficient of y to get a multiply of 4
$4(- \frac{1}{8})y= x^{2} -2.5x-7$
So the distance of the foci from the y-coordinate is - $\frac{1}{8}$ south to y-coordinate

Hence the coordinate of foci is (1.25, 17)

Function 4: following the steps above, the maximum value is when $x=8.5$ and $y=79.25$ . The distance from y-coordinate is 0.25 to the south of y-coordinate, hence the coordinate of foci is (8.5, 79.25-0.25)=(8.5,79)

Function 5: the minimum value of the function is when $x=-2.75$ and $y=-10.125$ . Manipulating coefficient of y, the distance of foci from y-coordinate is $\frac{1}{8}$ to the north. Hence the coordinate of the foci is (-2.75, -10.125+0.125)=(-2.75, -10)

Function 6: The maximum value happens when $x=1.5$ and $y=9.5$ . The distance of the foci from the y-coordinate is $\frac{1}{8}$ to the south. Hence the coordinate of foci is (1.5, 9.5-0.125)=(1.5, 9.375)

8 0

3 years ago

Stephanie's school is selling tickets to a play. On the first day of ticket sales the school sold 4 adult tickets and 10 student

klemol [59]

Answer:

The price of

1 adult ticket = $15

1 student ticket = $9

Step-by-step explanation:

Let

The price of adult tickets be represented by a

The price of student tickets be represented by s

Therefore:

On the first day of ticket sales the school sold 4 adult tickets and 10 student tickets for a total of $150.

4a + 10s = $150.... Equation 1

The school took in $105 on the second day by selling 1 adult ticket and 10 student tickets.

a + 10s = $105.... Equation 2

a = $105 - 10s

Therefore, we substitute : $105 - 10s = a in Equation 1

4a + 10s = $150.... Equation 1

4($105 - 10s) + 10s = $150

$420 - 40s + 10s = $150

Collect like terms

- 40s + 10s = $150 - $420

-30s = -$270

Divide both sides by -30

-30s/-30 = -$270/-30

s = $9

We find a

a = $105 - 10s

a = $105 - 10($9)

a = $105 - $90

a = $15

Therefore, the price of

1 adult ticket = $15

1 student ticket = $9

7 0

3 years ago

Is this correct? If not whats the correct answer

Illusion [34]

Yes I took the test

4 0

3 years ago

Read 2 more answers