The product of two consecutive integers is 342. What quadratic equation can be used to find x, the grater number?

Can someone please explain how the answer is D

Kitty [74]

Step-by-step explanation:

Because the length of ST is calculated using pythagorean theorem:

${c}^{2} = {a}^{2} + {b}^{2}$

Where the square of the hypotenuse is equal to the sum of squares of the other two sides of a right triangle. In this case, the hypotenuse is ST and the other two sides are distances between S and T over the X and Y axis. Those are easily calculated:

$x = |sx - tx| \\ y = |sy - ty|$

Where x is the distance between S and T over X axis and Y distance over Y axis, sx and tx are X coordinates of S and T, sy and ty are Y coordinates of S and T.

Using that formula, you get that y = 17 and x = 8.

Back to the pythagorean theorem, if we put those number in the formula of the pythagorean theorem, we get something like this:

${c}^{2} = {x}^{2} + {y}^{2} \\ {c}^{2} = {8}^{2} + {17}^{2} = 64 + 289 \\ {c}^{2} = 353 \\ c = \sqrt{353}$

And finally, the correct answer is in fact 353.

3 0

3 years ago

Two integers, n and p, have a product of -18. What is the largest possible sum of n and p?

Naily [24]

Answer:

17

Step-by-step explanation:

The two integers could be -1 and 18.

18 + -1 = 17

Any other combination of integers that have a product of -18 would have a smaller sum.

4 0

4 years ago

Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0

4 years ago

Given A(3, 4), B(4, 3) and C(2, 1), what are the coordinates of C', the image of C under a rotation of 90 degrees about the orig

Aleks [24]

<u>Given</u>:

The coordinates of the points A, B and C are (3,4), (4,3) and (2,1)

The points are rotated 90° about the origin.

We need to determine the coordinates of the point C'.

<u>Coordinates of the point C':</u>

The general rule to rotate the point 90° about the origin is given by

$(x, y) \Rightarrow(-y, x)$

Substituting the coordinates of the point C in the above formula, we get;

$(2, 1) \Rightarrow(-1, 2)$

Therefore, the coordinates of the point C' is (-1,2)

5 0

4 years ago

Irene tiene una colección de 50 dvd de películas de 90 minutos de duración cada una .Si el precio de cada uno era de 11€ , ¿cuan

Rainbow [258]

To solve this problem you must apply the proccedure shown below:

1. She has a total of 50 DVDs of 90 minutes each one of them. The cost of each DVD was 11€.

2. Therefore, to calculate the total cost of the collection, you must multiply the cost of each DVD by the total number of them:

$total=(11)(50)=550$ €

3. To calculate the total minutes of the collection, you must multiply 90 minutes by the total number of DVDs:

$(90min)(50)=4500min$

Therefore, the answer is: 550€ and 4500 minutes.

3 0

4 years ago