All categories

4 years ago

What is the solution of sqrt x+2-15=-3

Mathematics

1 answer:

antiseptic1488 [7]4 years ago

3 0

Answer:

Step-by-step explanation:

Without brackets, we are not exactly sure what is under the root sign. There are 3 choices.

sqrt(x) + 2 - 15 = - 3

sqrt(x + 2) - 15 = - 3

sqrt(x + 2 - 15) = - 3

I think the middle one is what you intend. If not leave a note.

sqrt(x + 2) - 15 = - 3 Add 15 to both sides.

sqrt(x + 2) - 15+15 = - 3+15 Combine

sqrt(x + 2) = 12 Square both sides

x + 2 = 12^2 Do the right

x + 2 = 144 Subtract 2 from both sides.

x + 2-2 = 144-2

x = 142

You might be interested in

(10 POINTS) answer ASAP if you can

Licemer1 [7]

Answer:

Step-by-step explanation:

A P E X

4 0

3 years ago

(5x10^3)x(9x10^7) can someone please answer this for me.

Anna11 [10]

4.5 times 10^10

1. Multiply 5times ten to the third

2. Multiply 9 times ten raised to the seventh

Then multiply them together
5000 times 90000000

5 0

4 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.

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