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3 years ago

102 over 250 as a decimal

Mathematics

1 answer:

natka813 [3]3 years ago

3 0

102/250 So, fractions basically mean division. So, you divide 102 by 250 and you get = 0.408 Answer = 0.408

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If a1 = 8 and an = an-1 + 5 then find the value of a5.

Nimfa-mama [501]

Step-by-step explanation:

A1=8

A2=A1+5 plug 8 into A1 here. After getting the value, put into next equation. And repeat until you get A5

A3=A2+5

A4=A3+5

A5=A4+5

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3 years ago

the bike is on sale for $130. if the original price of the bike $200, what is the percent discount ?

STALIN [3.7K]

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3 years ago

4. 4c - 3 (c = -2)<br><br><br> will give 5 stars

Flauer [41]

4(-2)-3 = -8-3 = -11
Answer: -11

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3 years ago

three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal

Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

= x + x + (x + $\frac{6}{5}$ ) = $\frac{27}{4}$

= 3x + $\frac{6}{5}$ = $\frac{27}{4}$

= 3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

= 3x = $\frac{135}{20}$ - $\frac{24}{20}$

= 3x = $\frac{111}{20}$

= x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag = $\frac{37}{20}$ kg

2nd bag = $\frac{37}{20}$ kg

3rd bag = $\frac{61}{20}$ kg

Hope it helps...

7 0

3 years ago

If possible help me in this. I need to find the two odd numbers...

Lelechka [254]

Part (a)

Consecutive odd integers are integers that odd and they follow one right after another. If x is odd, then x+2 is the next odd integer

For example, if x = 7, then x+2 = 9 is right after.

<h3>Answer: x+2</h3>

========================================================

Part (b)

The consecutive odd integers we're dealing with are x and x+2.

Their squares are x^2 and (x+2)^2, and these squares add to 394.

<h3>Answer: x^2 + (x+2)^2 = 394</h3>

========================================================

Part (c)

We'll solve the equation we just set up.

x^2 + (x+2)^2 = 394

x^2 + x^2 + 4x + 4 = 394

2x^2+4x+4-394 = 0

2x^2+4x-390 = 0

2(x^2 + 2x - 195) = 0

x^2 + 2x - 195 = 0

You could factor this, but the quadratic formula avoids trial and error.

Use a = 1, b = 2, c = -195 in the quadratic formula.

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(2)\pm\sqrt{(2)^2-4(1)(-195)}}{2(1)}\\\\x = \frac{-2\pm\sqrt{784}}{2}\\\\x = \frac{-2\pm28}{2}\\\\x = \frac{-2+28}{2} \ \text{ or } \ x = \frac{-2-28}{2}\\\\x = \frac{26}{2} \ \text{ or } \ x = \frac{-30}{2}\\\\x = 13 \ \text{ or } \ x = -15\\\\$

If x = 13, then x+2 = 13+2 = 15

Then note how x^2 + (x+2)^2 = 13^2 + 15^2 = 169 + 225 = 394

Or we could have x = -15 which leads to x+2 = -15+2 = -13

So, x^2 + (x+2)^2 = (-15)^2 + (-13)^2 = 225 + 169 = 394

We get the same thing either way.

<h3>Answer: Either 13, 15 or -15, -13</h3>

4 0

3 years ago