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4 years ago

The legs of a right triangle measure 4 and 6 inches. The area of the triangle is

Mathematics

2 answers:

leonid [27]4 years ago

8 0

Answer:

Area of triangle is 12 in.²

Step-by-step explanation:

we are given that:

The legs of a right triangle measure 4 and 6 inches.

(i.e. height and base of right angled triangle)

We have to find the area of this triangle.

area of triangle= $\dfrac{1}{2}\times base\times height$

= $\dfrac{1}{2}\times 4\times 6$

= 12 in.²

Hence, Area of triangle is 12 in.²

liraira [26]4 years ago

7 0

Area of a right triangle = .5 * leg1 * leg2
Area = .5* 24
Area = 12 square inches

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six times jason collection of books and one third of nathan collection add up to 134 books. one third of jason collection and na

raketka [301]

Answer:

j = 21 and n = 14

Step-by-step explanation:

we have the equations:

6j + n/3 = 134

j/3 + n = 31

54j + 3n = 1206

j + 3n = 93

53j = 1113

j = 21

(21)/3 + n = 31

7 + n = 31

n = 14

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4 years ago

Evaluate the expression x + 8/27. Evaluate the expression for x=8/9

diamong [38]

Answer:

32/27

Step-by-step explanation:

Evaluate for x=

8 /9 + 8 /27

=32/27

(Decimal: 1.185185)

6 0

3 years ago

Read 2 more answers

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0

3 years ago

Why will percent of change always be represented by a positive number?

Murljashka [212]

Percent of change will always be positive because the change is an absolute value. As we know, absolute value is always a positive value because you don't pay attention to the direction, only the distance is important.

3 0

3 years ago

List and describe the characteristics of a wave

o-na [289]

Answer:

Crest = Highest point of the wave.

Trough = Lowest point of the wave.

Wavelength = Distance from one crest/trough to the next (m)

Wave Height = Height from trough to crest (m)

Wave steepness = ratio of wave height to wavelength.

Amplitude = distance from the centre of wave to the bottom of the trough (m)

Step-by-step explanation:

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4 years ago