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3 years ago

6a+4=61/3+4 I don’t know what I’m doing wrong because I keep getting 10 1/3

Mathematics

1 answer:

Vlad1618 [11]3 years ago

4 0

Answer:

a = 3 7/18 or 3.388889

Step-by-step explanation:

First, you want to get the a by itself, so you would subtract 4 from the left side of the equals sign.

6a + 4 = 61 / 3 + 4

-4 -4

6a = 61 / 3

By doing that, they cancel each other out, so you are left with

6a = 61 / 3

Then you want to get a by itself, so you divide 6 from both sides.

6a / 6 = 61 / 3 / 6

Remember that 6 can also be written as 6/1

Remember that dividing a fraction is multiplying the reciprocal.

6a / 6 = 61 / 3 / 6 / 1

When dividing fractions remember to Skip the first fraction, Flip the second fraction, and Multiply the two fractions together.

61 / 3 x 1 / 6

skip multiply flip

Multiply across the top, and across the bottom.

61(1) and 3(6)

Then you get

61 / 18

So a = 61 / 18

If you want to convert that to a mixed number, it would be 3 7 / 18

If you want that in a decimal, it would be 3.388889

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Maci and I are making a small kite. Two sides are 10". Two sides are 5". The shorter diagonal is 6". Round all your answers to t

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Answer:

A. 4".

B. Approximately 9.54".

C. Approximately 13.54".

Step-by-step explanation:

Please find the attachment.

Let x be the distance from the peak of the kite to the intersection of the diagonals and y be the distance from the peak of the kite to the intersection of the diagonals.

We have been given that two sides of a kite are 10 inches and two sides are 5 inches. The shorter diagonal is 6 inches.

A. Since we know that the diagonals of a kite are perpendicular and one diagonal (the main diagonal) is the perpendicular bisector of the shorter diagonal.

We can see from our attachment that point O is the intersection of both diagonals. In triangle AOD the side length AD will be hypotenuse and side length DO will be one leg.

We can find the value of x using Pythagorean theorem as:

$(AO)^2=(AD)^2-(DO)^2$

$x^{2}=5^2-3^2$

$x^{2}=25-9$

$x^{2}=16$

Upon taking square root of both sides of our equation we will get,

$x=\sqrt{16}$

$x=\pm 4$

Since distance can not be negative, therefore, the distance from the peak of the kite to the intersection of the diagonals is 4 inches.

B. We can see from our attachment that point O is the intersection of both diagonals. In triangle DOC the side length DC will be hypotenuse and side length DO will be one leg.

We can find the value of y using Pythagorean theorem as:

$(OC)^2=(DC)^2-(DO)^2$

Upon substituting our given values we will get,

$y^2=10^2-3^2$

$y^2=100-9$

$y^2=91$

Upon taking square root of both sides of our equation we will get,

$y=\sqrt{91}$

$y\pm 9.539392$

$y\pm\approx 9.54$

Since distance can not be negative, therefore, the distance from intersection of the diagonals to the top of the tail is approximately 9.54 inches.

C. We can see from our diagram that the length of longer diagram will be the sum of x and y.

$\text{The length of the longer diagonal}=x+y$