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Alexxandr [17]
3 years ago
5

Is the square root of any positive number irrational

Mathematics
1 answer:
ser-zykov [4K]3 years ago
5 0
No because if it is the square root of 100, then it is rational, hope this helps
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Can someone please factor this expression for me a²+4a-45​
NARA [144]

Answer:

(a + 9)(a - 5)

Step-by-step explanation:

Given

a² + 4a - 45

Consider the factors of the constant term (- 45) which sum to give the coefficient of the a- term (+ 4)

The factors are + 9 and - 5 , since

9 × - 5 = - 45 and 9 - 5 = 4 , then

a² + 4a - 45 = (a + 9)(a - 5) ← in factored form

8 0
3 years ago
You have 600 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river
inn [45]

Answer:

The length and width of plot is L=300\:ft, W=150\:ft

Largest area of the plot is A=45000 ft^{2}

Step-by-step explanation:

Assume width as x and length as y. Given that length of fencing is 600 feet and fencing is enclosed on 3 sides. So perimeter is given as,

Perimeter = width + length + width

Substituting the value,  

600=x+y+x

600= 2x+y  ….1

Now area of fence of rectangular box is given as follows,  

A=xy  ….2

Solving equation 1 for y, subtracting 2x from both sides,  

600-2x= y

Substituting the value in equation 2,  

A=x\left (600-2x  \right )

Simplifying

A=600x-2x^{2}

Rewriting,  

A=-2x^{2}+600x

Above equation looks like quadratic equation f\left ( x \right )=ax^{2}+bx+c whose graph looks like parabola.  

Comparing equation f(x) and A values of a, b and c are, a=-2,b=600 and c=0.  

Now maximum of f\left ( x \right ) occurs at vertex.  

The x coordinate of the vertex is given as -\dfrac{b}{2a}

Substituting the values,  

x=-\dfrac{600}{2\left (-2  \right )}

Simplifying,  

x=\dfrac{600}{4}

x=150

So width of plot is 150 feet.  

Now to calculate value of length by using equation 1,

600= 2x+y

Substituting the values,

600= 2\left (150  \right )+y

600= 300+y

Subtracting 300 from both sides,

300= y

So length of plot is 300 ft.

The y coordinate of the vertex is given as y=f\left ( -\dfrac{b}{2a} \right ) which also means, y=f\left ( 150 \right )

\therefore A=-2\left (150  \right )^{2}+600\left (150  \right )

Simplifying,  

\therefore A=-2\left (22500  \right )^{2}+600\left (150  \right )

\therefore A=-45000+90000

\therefore A=45000

So, area of the plot will be A=45000 ft^{2}

7 0
3 years ago
I NEED HELP PLEASE !!!!
Darya [45]

Answer:

a and c

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
1 gallon minus 1 quart 1 pint and 1 ounce?
barxatty [35]
That would be zero becuase there are 4 quarts ina gallon and 8 pnts and 16 ounces
5 0
3 years ago
Read 2 more answers
I need help with questions #7 and #8 plz
katen-ka-za [31]

Answer:

7. A = 40.8 deg; B = 60.6 deg; C = 78.6 deg

8. A = 20.7 deg; B = 127.2 deg; C = 32.1 deg

Step-by-step explanation:

Law of Cosines

c^2 = a^2 + b^2 - 2ab \cos C

You know the lengths of the sides, so you know a, b, and c. You can use the law of cosines to find C, the measure of angle C.

Then you can use the law of cosines again for each of the other angles. An easier way to solve for angles A and B is, after solving for C with the law of cosines, solve for either A or B with the law of sines and solve for the last angle by the fact that the sum of the measures of the angles of a triangle is 180 deg.

7.

We use the law of cosines to find C.

18^2 = 12^2 + 16^2 - 2(12)(16) \cos C

324 = 144 + 256 - 384 \cos C

-384 \cos C = -76

\cos C = 0.2

C = \cos^{-1} 0.2

C = 78.6^\circ

Now we use the law of sines to find angle A.

Law of Sines

\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}

We know c and C. We can solve for a.

\dfrac{a}{\sin A} = \dfrac{c}{\sin C}

\dfrac{12}{\sin A} = \dfrac{18}{\sin 78.6^\circ}

Cross multiply.

18 \sin A = 12 \sin 78.6^\circ

\sin A = \dfrac{12 \sin 78.6^\circ}{18}

\sin A = 0.6535

A = \sin^{-1} 0.6535

A = 40.8^\circ

To find B, we use

m<A + m<B + m<C = 180

40.8 + m<B + 78.6 = 180

m<B = 60.6 deg

8.

I'll use the law of cosines 3 times here to solve for all the angles.

Law of Cosines

a^2 = b^2 + c^2 - 2bc \cos A

b^2 = a^2 + c^2 - 2ac \cos B

c^2 = a^2 + b^2 - 2ab \cos C

Find angle A:

a^2 = b^2 + c^2 - 2bc \cos A

8^2 = 18^2 + 12^2 - 2(18)(12) \cos A

64 = 468 - 432 \cos A

\cos A = 0.9352

A = 20.7^\circ

Find angle B:

b^2 = a^2 + c^2 - 2ac \cos B

18^2 = 8^2 + 12^2 - 2(8)(12) \cos B

324 = 208 - 192 \cos A

\cos B = -0.6042

B = 127.2^\circ

Find angle C:

c^2 = a^2 + b^2 - 2ab \cos C

12^2 = 8^2 + 18^2 - 2(8)(18) \cos B

144 = 388 - 288 \cos A

\cos C = 0.8472

C = 32.1^\circ

8 0
3 years ago
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