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2 years ago

Simplify 6^7/6^5 I have no idea math is hard

Mathematics

2 answers:

MrMuchimi2 years ago

8 0

Step-by-step explanation:

hi hlo hshsvbzbzbsbbzbsbsbsb

Delicious77 [7]2 years ago

5 0

Answer:

It is 36

Step-by-step explanation:

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if alpha and beta are zeroes of the quadratic polynomial f(x) = x2-x-2 then find a polynomial whose zeroes are 2alpha + 1 and 2b

frosja888 [35]

Answer:

Step-by-step explanation:

Hello, as alpha and beta are zeroes of

$x^2-x-2$

it means that their sum is alpha+beta=1 and their product alpha*beta=-2.

The polynomial whose zeroes are 2 alpha + 1 and 2 beta + 1, means that the sum of its zeroes is 2(alpha+beta)+2=2+2=4

and the product is (2alpha+1)(2beta+1)=4 alpha*beta + 2(alpha+beta) + 1 = 4 * (-2) + 2*(1) +1 = -8 + 2 + 1 = -5. so one of these polynomials is

$\Large \boxed{\sf \bf \ \ x^2-4x-5 \ \ }$

Thank you.

5 0

2 years ago

What is the measure of А 50" 50° B C [Not drawn to scale] 50° 80 100 130

Paladinen [302]

Answer:

solution given :

<A=?

<B=50°

<C=50°

if it is a triangle:

<A+<B+<C=180°( sum of interior angle of a triangle is 180°)

<A=180-50-50

<A=80°

<A=80° is a required answer.

7 0

2 years ago

Question 17 of 33

butalik [34]

Answer:

Step-by-step explanation:

i took the test

8 0

2 years ago

I(5, 3), J(5, -3), and L(-4, 3) are three vertices of rectangle IJKL. What are the coordinates of the fourth vertex, K, of recta

Alex_Xolod [135]

4,-3 maybe idk for sure

7 0

2 years ago

Read 2 more answers

Two vertical poles of length 10 and 12 feet, respectively, stand 17 feet apart. A cable reaches from the top of one pole to the

Marat540 [252]

Answer:

$L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433} ft$

Step-by-step explanation:

let length of calble = L

distance from 10-foot pole = x ft

Moving from the top of the 10-foot pole to point x, we have

From the diagram that

$x^{2} +10^{2}=L_{1} ^{2} \\x^{2} + 100 = L_{1} ^{2}$

Take the square root of both sides

$\sqrt{x^{2}+100}=\sqrt{L_{1} ^{2}}\\ \sqrt{x^{2}+100}= L_{1}$

Moving from point x to the top of the 12-foot pole, we have

From the diagram that

$(17-x)^{2}+12^{2}=L_{2} ^{2} \\289-34x+x^{2} +144=L_{2} ^{2}\\x^{2} -34x+433=L_{2} ^{2}$

Take the square root of both sides

$\sqrt{x^{2}-34x+433}=\sqrt{L_{2} ^{2}} \\\sqrt{x^{2}-34x+433}=L_{2}$

Amount of cable used, L = L1 + L2

$L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433}$

3 0

2 years ago