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2 years ago

Help!! I need help!

Mathematics

1 answer:

Tasya [4]2 years ago

6 0

Answer:

Step-by-step explanation:

-5x-49>113

+49 +49

———————-

-5x>162

—- ——

-5 -5

x<32.4

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(no links or files) Solve this equation using the quadratic formula. This question should be solved both ways. (1) Quadratic For

meriva

Quadratic formula is x = -b+ or- sq rt b^2-4ac / 2a
a=2 b=5 c=-3
-5 +or- sqrt 5^2-4(2)(-3) / 2(2)
-5 +or- sqrt 49/ 4
-5 + 7 /4 = 2/4 = 1/2
-5 - 7 /4 = -12/4 = -3

Factoring a*c is 2*-3 =-6
Factors of -6 that add to 5 are 6 and -1
Split 5x into +6x-1x
2x^2+6x-1x-3 and group
2x(x+3)-1(x+3)
(x+3)(2x-1)=0
x+3=0 gives x=-3
2x-1=0 gives x=1/2

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2 years ago

Which elements in a set below are integers?12,-3/4,5,-7,2.24,14/7,9

zhuklara [117]

Answer:

12, 5, -7, 14/7, 9

Step-by-step explanation:

Integers are numbers that are either positive or negative and are whole numbers (no decimals).

4 0

3 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

3 years ago

Convert 7.5 to a decimal

julsineya [31]

7.5 is already in decimal form.

3 0

3 years ago

Use technology and the given confidence level and sample data to find the confidence interval for the population mean mu. Assume

Sphinxa [80]

Answer: (1.55, 6.45)

Step-by-step explanation:

The confidence interval for population mean is given by :-