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dimulka [17.4K]
3 years ago
9

Solve the inequality 3x^2 + 10x - 8 < 0​

Mathematics
1 answer:
Ksivusya [100]3 years ago
8 0

Step-by-step explanation:

Given, 3x−2<2x+1

⇒3x−2x<1+2

⇒x<3orx∈(−∞,3)

The lines y=3x−2 and y=2x+1 both will intersect at x=3

Clearly, the dark line shows the solution of 3x−2<2x+1.

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X^3+3x^5=x5 what's the missing expression
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3 years ago
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Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

8 0
4 years ago
Hey please help me please help please
Airida [17]

1.

a) metres to centimetres :

multiply length by 100

b) metres to millimetres:

multiply length by 1000

c) kilograms to grams:

multiply the mass value by 1000

d) litres to millilitres :

multiply volume by 1000

2.

a) 3 m = 3× 100 = 300 cm

b) 28 cm = 28 × 10 = 280 mm

c) 2.4 km = 2.4 × 1000

= 24 × 10^-1 × 10^3

= 24 × 10^2 =2400 m

d) 485 mm =485 / 10

= 485 / 10 ^1

= 485 × 10 ^-1

= 48.5 cm

e) 35 cm = 35 / 100

= 35 /10^2

= 35 × 10 ^ -2

= 0.35 m

f) 2.4 m = 2.4 / 1000

= 24 × 10 ^-1 / 10^3

= 24 × 10^-1 × 10 ^-3

= 24 × 10 ^ -4

= 0.0024 km

g) 2495 mm = 2495 /1000

= 2495 /10^ 3

= 2495 × 10 ^-3

=2.495 m

4 0
3 years ago
4. Tikili wants to invest $3500 at her local bank. She will receive an interest rate of 5% compounded
alexandr402 [8]

Answer:

3863.35 = 3500(1 +  \frac{0.05}{2} )^{(2 \times 2)}

Step-by-step explanation:

The formula for this equation is

a = p(1 +  \frac{r}{n} )^{(n \times t)}

a is the final result

p is the starting amount (deposited)

r is the interest rate

n is the number of times it's compounded

t is the time

because it says compound annually and it's after 2 years both t and n equal 2. I rounded a for you, but if you don't need it rounded here it is: 3863.345117

Please double check me I may be wrong, this is my second time doing these type of questions

3 0
3 years ago
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