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3 years ago

10

Simplify and leave in radical form.

" id="TexFormula1" title="\sqrt{3^{3} \sqrt{2}" alt="\sqrt{3^{3} \sqrt{2}" align="absmiddle" class="latex-formula">

1 answer:

BaLLatris [955]3 years ago

4 0

Answer:

3 $\sqrt{3}$ $\sqrt[4]{2}$

Step-by-step explanation:

(3^3 * 2^(1/2))^1/2 = 3^(3/2) * 2^(1/4) = 3* 3^(1/2) * 2^(1/4)

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I’ll give brainliest

motikmotik

Answer:

v= 5/6

Step-by-step explanation:

You have to divide -6 on both sides to isolate the variable V. Then it would become -5/-6. The negative signs cancel out.

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Car A travels at 45 miles per hour and Car B travels at a greater speed of x miles per hour. Write and simplify an expression th

jonny [76]

$farther = 2 \times (x - 45) = 2x - 90$

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3 years ago

Give the values of a, b, and c needed to write the equation's standard form.(5 + x)(5 - x) = 7A = 1; B = 0; C = -18A = 25; B = 0

Maksim231197 [3]

We need to FOIL that out and then get it into standard form to see what our A, B, and C are. FOILing gives us $25-x^2=7$ . If we move the 7 over by subtraction and set it equal to 0, we have $-x^2+18=y$ . In order to avoid a negative leading coefficient, we can change all the signs to get $x^2-0x-18=0$ so A = 1, B = 0, C = -18, first choice above.

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2 years ago

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Ahmad should have $52 left

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Which is an equation of a circle with center (5, 0) that passes through the point (1, 1)

alex41 [277]

Equation of circle at center (h,k) is given by

$(x-h)^2+(y-k)^2=r^2$

Given that center is at (5,0) that means h=5 and k=0

plug both values into above formula

$(x-5)^2+(y-0)^2=r^2$

$(x-5)^2+y^2=r^2$ ...(i)

Given that circle passes through point (1,1) so it will satisfy above equation

$(1-5)^2+1^2=r^2$

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$16+1=r^2$

$17=r^2$

Now plug this value of r^2 into equation (i)

$(x-5)^2+y^2=17$

which best matches with choice C

Hence $(x-5)^2+y^2=17$ is the final answer.

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3 years ago