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Hatshy [7]
3 years ago
11

PLS HELP WILL MARK BRAINLIEST

Mathematics
1 answer:
MArishka [77]3 years ago
4 0

Answer:

3/7

Step-by-step explanation:

Third option is correct answer ig!

Hope it is the correct answer

Mark as brainliest so tht i can earn some points

Be humble enough to mark as barinloest because u wont loose money to mark me as brainliest

So plzz mark as barinliest!!!

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I know yaw no this stuff
Aleks04 [339]

Answer:

68

Step-by-step explanation:

too easy

7 0
3 years ago
An airline wants to select a computer software package for its reservation system. Four software packages (1, 2, 3, and 4) are c
zzz [600]

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4 0
4 years ago
Find the side indicated by the variable. Round to the nearest tenth.​
zepelin [54]

Answer:

b ≈ 25.6

Step-by-step explanation:

From the figure attached,

By applying tangent rule in the given triangle,

tan(32°) = \frac{\text{Opposite side}}{\text{Adjacent side}}

tan(32°) = \frac{16}{b}

b = \frac{16}{\text{tan}(32)}

b = \frac{16}{0.62487}

b = 25.605

b ≈ 25.6

5 0
3 years ago
Solve using the quadratic formula<br> 2x^2+x+67=0
telo118 [61]

Answer:

\displaystyle x=\frac{-1 \pm i\sqrt{535}}{2}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

  • Multiple Roots
  • Standard Form: ax² + bx + c = 0
  • Quadratic Formula: \displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

<u>Algebra II</u>

  • Imaginary Root <em>i</em> = √-1

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

2x² + x + 67 = 0

<em>a</em> = 2

<em>b</em> = 1

<em>c</em> = 67

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Substitute in variables [Quadratic Formula]:                                                  \displaystyle x=\frac{-1 \pm \sqrt{1^2-4(2)(67)}}{2(1)}
  2. Multiply:                                                                                                             \displaystyle x=\frac{-1 \pm \sqrt{1^2-4(2)(67)}}{2}
  3. [√Radical] Evaluate exponents:                                                                       \displaystyle x=\frac{-1 \pm \sqrt{1-4(2)(67)}}{2}
  4. [√Radical] Multiply:                                                                                           \displaystyle x=\frac{-1 \pm \sqrt{1-536}}{2}
  5. [√Radical] Subtract:                                                                                          \displaystyle x=\frac{-1 \pm \sqrt{-535}}{2}
  6. [√Radical] Simplify:                                                                                           \displaystyle x=\frac{-1 \pm i\sqrt{535}}{2}
3 0
3 years ago
Use mathematical induction to prove the statement is true for all positive integers n. 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = (n(2n-
Charra [1.4K]

Answer:

The statement is true is for any n\in \mathbb{N}.

Step-by-step explanation:

First, we check the identity for n = 1:

(2\cdot 1 - 1)^{2} = \frac{2\cdot (2\cdot 1 - 1)\cdot (2\cdot 1 + 1)}{3}

1 = \frac{1\cdot 1\cdot 3}{3}

1 = 1

The statement is true for n = 1.

Then, we have to check that identity is true for n = k+1, under the assumption that n = k is true:

(1^{2}+2^{2}+3^{2}+...+k^{2}) + [2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}

\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)}{3} +[2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}

\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot [2\cdot (k+1)-1]^{2}}{3} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}

k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot (2\cdot k +1)^{2} = (k+1)\cdot (2\cdot k +1)\cdot (2\cdot k +3)

(2\cdot k +1)\cdot [k\cdot (2\cdot k -1)+3\cdot (2\cdot k +1)] = (k+1) \cdot (2\cdot k +1)\cdot (2\cdot k +3)

k\cdot (2\cdot k - 1)+3\cdot (2\cdot k +1) = (k + 1)\cdot (2\cdot k +3)

2\cdot k^{2}+5\cdot k +3 = (k+1)\cdot (2\cdot k + 3)

(k+1)\cdot (2\cdot k + 3) = (k+1)\cdot (2\cdot k + 3)

Therefore, the statement is true for any n\in \mathbb{N}.

4 0
3 years ago
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