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2 years ago

In isosceles triangle WXY, XY is 13 inches in length and WY is 10 inches. What is the length of XZ? answer asap please

Mathematics

1 answer:

Anna11 [10]2 years ago

8 0

<h3>Answer: 12 inches</h3>

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Explanation:

Notice the double tickmarks on segments WZ and ZY. This tells us the two segments are the same length. Let's say they are m units long, where m is a placeholder for a positive number.

That would mean m+m = 2m represents the length of segment WY, but that's equal to 10 as the diagram shows. We have 2m = 10 lead to m = 5 after dividing both sides by 2.

We've shown that WZ and ZY are 5 units long each. In short, we just cut that length of 10 in half.

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Let's focus on triangle XYZ. This is a right triangle with legs XZ = unknown and ZY = 5. The hypotenuse is XY = 13.

We'll use the pythagorean theorem to find XZ

a^2 + b^2 = c^2

(XZ)^2 + (ZY)^2 = (XY)^2

(XZ)^2 + (5)^2 = (13)^2

(XZ)^2 + 25 = 169

(XZ)^2 = 169-25

(XZ)^2 = 144

XZ = sqrt(144)

XZ = 12

Segment XZ is 12 inches long.

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Which expression is equivalent to (-3)x+ (-3) (-12)?

julia-pushkina [17]

Answer:

Step-by-step explanation:

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2 years ago

How many numbers are between 1 and 3 ?

kramer

If we are talking about whole numbers, the only number in between 1 and 3 is 2 but if we are talking about numbers in general, there are an infinite amount of numbers (for example, 1.1, 1.12, 1.004, and 1.80543285 are all in between 1 and 3).

5 0

3 years ago

I really need help quick

Natali [406]

Answer:

$\sf x=2+\sqrt{6},\:x=2-\sqrt{6}$

Explanation:

$\sf y = x^2 -4x$

given that y = 2

$\sf x^2 -4x = 2$

$\sf x^2 -4x -2 = 0$

using quadratic formula:

$\sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \ \ \ when \ ax^2 + bx + c = 0$

$\rightarrow \sf x_{1,\:2}=\dfrac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:1\cdot \left(-2\right)}}{2\cdot \:1}$

$\rightarrow \sf x_1=\dfrac{-\left(-4\right)+2\sqrt{6}}{2\cdot \:1},\:x_2=\dfrac{-\left(-4\right)-2\sqrt{6}}{2\cdot \:1}$

$\rightarrow \sf x=2+\sqrt{6},\:x=2-\sqrt{6}$

in decimals:

$\rightarrow \sf x=4.45 , \ -0.45$

5 0

2 years ago

Length of a rectangle is 5 cm longer than the width. Four squares are constructed outside the rectangle such that each of the sq

stiv31 [10]

Answer:

Step-by-step explanation:

Remark

This is one of those questions that can throw you. The problem is that do you include the original rectangle or not. The way it is written it sounds like you shouldn't

However if you don't the question gives you 2 complex answers. (answers with the sqrt( - 1) in them.

Solution

Let the width = x

Let the length = x + 5

Area of the rectangle: L * w = x * (x + 5)

Area of the smaller squares (there are 2)

Area = 2*s^2

x = s

Area = 2 * x^2

Area of the larger squares = 2 * (x+5)^2

Total Area

x*(x + 5) + 2x^2 + 2(x + 5)^2 = 120 Expand

x^2 + 5x + 2x^2 + 2(x^2 + 10x + 25) = 120 Remove the brackets

x^2 + 5x + 2x^2 + 2x^2 + 20x + 50 = 120 collect the like terms on the left

5x^2 + 25x + 50 = 120 Subtract 120 from both sides.

5x^2 + 25x - 70 = 0 Divide through by 5

x^2 + 5x - 14 = 0 Factor

(x + 7)(x - 2) = 0 x + 7 has no meaning

x - 2 = 0

x = 2

Perimeter

P = 2*w + 2*L

w = 2

L = 2 + 5

L = 7

P = 2*2 + 2 * 7

P = 4 + 14

P = 18

8 0

2 years ago

In 2010, the number of houses built in Town A was 25 percent greater than the number of houses built in Town B. If 70 houses wer

Stella [2.4K]

Answer:

The number of houses built in Town B is 56.

Step-by-step explanation:

We are given that in 2010, the number of houses built in Town A was 25 percent greater than the number of houses built in Town B.

Also, 70 houses were built in Town A during 210.

Let the number of houses built in Town B be 'x'.

So, according to the question;

Number of houses built in Town A = Number of houses built in Town B + 25% of the houses built in Town B

$70 = x + (25\% \times x)$

$70 = x(1+0.25)$

$x=\frac{70}{1.25}$

x = 56

Hence, the number of houses built in Town B is 56.

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2 years ago