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1 year ago

WILL GIVE BRAINLIEST

Mathematics

1 answer:

vagabundo [1.1K]1 year ago

8 0

Answer:

It has no solution because, 14x cancels out -14x when you move the variables to the left and leaves the equation as 0 = -12, however we know that 0 does not equal -12 which means this solution is false; therefore this equation has no solution since 0 does not equal -12.

Step-by-step explanation:

5 - 4x - 7 - 10x = -8x - 4 - 6x - 10 (Given)

-2 - 14x = -14x - 14 (Combine like terms)

0 = -12 (Move variables(x) to the left and the constants to the right)

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There is a circle with a center of 0,0 on a coordinate plane. There is one point on the circle's circumference in which the x:y

abruzzese [7]

Answer:

Step-by-step explanation:

Suppose the radius is 1. The parametric equations for the circle are

x = cosθ

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The solutions are (3/√10, 1/√10) and (-3/√10, -1/√10).

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2 years ago

Whats the difference between coplanar and collinear

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2 years ago

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A is 60% of B , B is 30% of C, what percentage of C is A

zaharov [31]

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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...

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