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3 years ago

I need help with number 3 please help it’s due on nov 4th.

Mathematics

1 answer:

Kitty [74]3 years ago

6 0

I’ll explains in the best way I can. Imagine the 3rd one but on the bottom square add one on the left and then one under that. So you would add a total of 2 squares. Hope this helps :)

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Help please thank you

Ad libitum [116K]

In order for the triangle to be isosceles, we have to set two lengths of the triangle equal to each other.

Let's take the lengths 5x-12 and x+20 and set them equal to each other.

5x - 12 = x + 20

Combine like terms by moving them over to their respective sides.

Subtract x from both sides of the equation.

4x - 12 = 20

Add 12 to both sides of the equation.

4x = 32

Divide both sides by 4.

x = 8

Check your answer by substituting.

5x - 12 = x + 20

5(8) - 12 = 8 + 20

40 - 12 = 28

28 = 28

Solution: x = 8

5 0

3 years ago

Which is an algebraic expression for five more then Z?

faltersainse [42]

I believe the answer would be Z < (Z+5)

4 0

3 years ago

Read 2 more answers

Why do the inequality signs stay the same in the last two steps of exercise 1

MrRissso [65]

The only reason an inequality would change would be because a number is divided or multiplied by a negative number. So if it stayed the same, then it would be because there was no division or multiplication by a negative number. Hope I helped :)

7 0

3 years ago

Read 2 more answers

What are the next three terms in the pattern 2, 6, 18, 54, ...?

castortr0y [4]

Answer:

OPTION B: 162, 486, 1458

Step-by-step explanation:

The given sequence is 2, 6, 18, 54, . . .

It is a geometric sequence and the common difference is 3.

The general form of a geometric sequence is: a, ar, ar², ar³, . . .

Here a = 2 and r = 3.

$n^{th}$ term of a Geometric progression is $ar^{n - 1}$ .

Note that the fourth term is 54.

i.e., $ar^3 = 54$

$\implies ar^4 = ar^3 . r = 54 . 3 = 162$ .

Similarly, $ar^6 = 162 \times 3 = 486$ .

Also, $ar^7 = ar^6 . r = 486 \times 3 = 1458$ .

Hence, OPTION B is the answer.

8 0

3 years ago

Read 2 more answers

Find a polynomial function of least degree having only real coefficients, a leading coefficient of 1, and roots of 2-√6 , 2+ √6

horsena [70]

Answer:

$x^{4} -18x^{3}+104x^{2} -172x-100$

Step-by-step explanation:

The 3 roots are given out of which 2 are real and 1 is imaginary. For a polynomial of least degree having real coefficients, it must have a complex conjugate root as the 4th root. Therefore, based on 4 roots, the least degree of polynomial will be 4. Finding the polynomial having leading coefficient=1 and solving it based on multiplication of 2 quadratic polynomials, we get:

$\\\\x_{1} = 2-\sqrt{6} \\x_{2} = 2+\sqrt{6} \\x_{3}=7-i \\x_{4}=7+i \\\\P(x)=1(x-x_{1})(x-x_{2} )(x-x_{3} )(x-x_{4} ) \\\\=(x-(2-\sqrt{6}))( x-(2+\sqrt{6} )) (x-(7-i))( x-(7+i))\\=((x-2)+\sqrt{6})( ( x-2)-\sqrt{6} ) ((x-7)+i)( (x-7)-i)\\=((x-2)^{2} -(\sqrt{6} )^{2} )((x-7)^{2}-(i)^{2})\\=(x^{2} -4x-2)(x^{2} -14x+50)\\=x^{4} -18x^{3}+104x^{2} -172x-100\\$

7 0

3 years ago