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3 years ago

Evaluate the variable expression x + y + z for x = –17.17, y = 34, and z = –3.9264.

Mathematics

1 answer:

lapo4ka [179]3 years ago

6 0

The answer is
12.9036

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Find 3 consecutive integers such that the sum of the first and twice the second is 16 more than twice the third

lions [1.4K]

Answer: 18, 19, 20

Step-by-step explanation:

3 consecutive integers = x, y and z

1st number is X

2nd number is Y which is X +1

3rd number is Z which is X +2

Write equation

X + 2Y = 16 + 2Z

Replace Y with X + 1 and Z with X + 2

So,

X + 2(X + 1) = 16 + 2(X + 2)

Distribute

X + 2X + 2 = 16 + 2X + 4

Combine like terms

3X + 2 = 2X + 20

Solve for X

3X + 2 - 2 - 2X = 20 + 2X - 2X - 2

X = 18 so, Y is 19 and Z is 20

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3 years ago

PLS I BEG! Help! Fractions! Put the fraction in order from least to greatest. Il lgive brainlist!

baherus [9]

Answer:

2a. 4/9 < 1/2 < 4/4

3a. 4/9 < 1/2 < 5/4

Step-by-step explanation:

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3 years ago

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Michael has saved $22.40. Kevin saved 3/4 of the amount that Michael saved. Jack saved 5 times as much as Kevin. What is the tot

Schach [20]

202.5 is the answer

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3 years ago

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In the game of "odd man out" each player tosses a fair coin. if all the coins turn up the same except for one, the player tossin

Temka [501]

The options are H,H,T or T,T,H

HHT: 1/2 x 1/2 x 1/2 = 1/8

TTH: 1/2 x 1/2 x 1/2 = 1/8

HHT or TTH: 1/8 + 1/8 = 2/8 = 1/4

Answer: 1 out of 4 = 1/4 = 25%

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3 years ago

If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?

lys-0071 [83]

Answer:

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

$3x^2-(3k-2)x-(k-6) \text{ with }k=2$ :

$3x^2-(3\cdot 2-2)x-(2-6)$

$3x^2-4x+4$

I'm going to solve $3x^2-4x+4=0$ for x using the quadratic formula:

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}$

$\frac{4\pm \sqrt{16-16(3)}}{6}$

$\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}$

$\frac{4\pm 4\sqrt{-2}}{6}$

$\frac{2\pm 2\sqrt{-2}}{3}$

$\frac{2\pm 2i\sqrt{2}}{3}$

Let's see if uv=u+v holds.

$uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}$

Keep in mind you are multiplying conjugates:

$uv=\frac{1}{9}(4-4i^2(2))$

$uv=\frac{1}{9}(4+4(2))$

$uv=\frac{12}{9}=\frac{4}{3}$

Let's see what u+v is now:

$u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}$

$u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$

We have confirmed uv=u+v for k=2.

4 0

3 years ago