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3 years ago

WILL GIVE BRAINLST HAVE A GOOD DAY

Mathematics

2 answers:

lorasvet [3.4K]3 years ago

8 0

National Kala KS so Liz

nikitadnepr [17]3 years ago

8 0

Answer:

13)

$\sqrt{44} \\= 2\sqrt{11}$

14)

Step-by-step explanation:

13) If you you remove 4 from 44, you have to find the square root of 4, which is 2 then you multiply that by whatever is left inside.

14) Rational

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You find yourself with 17 guilders( a gold coin) but you are planning on taking a trip to a country where the local currency is

3241004551 [841]

We are given

The current exchange rate is 2 guilders for every 3 florins

so,

2 guilders = 3 florins

now, we can find for 1 guilders

so, we get

$1guilders=\frac{3}{2} florins$

now, we can multiply both sides by 17

$17\times 1guilders=17\times \frac{3}{2} florins$

now, we can simplify it

$17guilders=25.5florins$

so, option-A........Answer

5 0

3 years ago

Solving simultaneous equations with one quadratic, x^2+y^2=16 and y=x+3

Svetlanka [38]

If $y=x+3$ , then in the first equation you have

$x^2+(x+3)^2=x^2+(x^2+6x+9)=2x^2+6x+9=16$

Rewriting a bit, you get

$x^2+3x+\dfrac92=8$

$x^2+3x+\dfrac94+\dfrac94=8$

$\left(x+\dfrac32\right)^2=\dfrac{23}4$

$x=-\dfrac32\pm\dfrac{\sqrt{23}}2$

Now, since $y=x+3$ , you also get

$y=-\dfrac32\pm\dfrac{\sqrt{23}}2+3=\dfrac32\pm\dfrac{\sqrt{23}}2$

So, there are two solutions here:

$(x,y)=\left(-\dfrac32+\dfrac{\sqrt{23}}2,\dfrac32+\dfrac{\sqrt{23}}2\right)$
$(x,y)=\left(-\dfrac32-\dfrac{\sqrt{23}}2,\dfrac32-\dfrac{\sqrt{23}}2\right)$

7 0

3 years ago

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.

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