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3 years ago

What is 328 divided by 23 with the remainder

Mathematics

2 answers:

Nata [24]3 years ago

8 0

Answer:

7.1

Step-by-step explanation:

328 23 = 14.2608695652

14.2608695652 = 142.608695652 to the nearest tenth

14.2608695652 = 14.26 to the nearest hundredth

14.2608695652 = 14.261 to the nearest thousandth

= 0 to the nearest tenth

= 0 to the nearest hundredth

= 0 to the nearest thousandth

Other Divisions Math homework are

328 divide by half plus 20

Homework answers: (328/2) + 20 = 184

328 divide by half plus 40

Homework answers: (328/2) + 40 = 204

328/23 divided by 2

Answer: (328/23) ÷ 2 = 7.13043478261

Andrews [41]3 years ago

6 0

14.27 if rounded up if not den 14.26086956521739

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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

Art [367]

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.

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