Find the area of the triangle 19cm 15cm 12cm

Mathematics

1 answer:

777dan777 [17]3 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

The coordinates of trapezoid ABCD are A(−4, 3), B(2, 3), C(4, −1) and D(−4, −1). A line segment runs through the trapezoid with

madam [21]

The distance from point Y to the y-axis is 4 units and the distance from point Z to the y-axis is 3 units, then the lelgth of the segment YZ is 4+3=7 units.
If a scale factor is 3, then the length of Y'Z' will be 7·3=21 units.

P.S. In the added picture you can see trapezoid ABCD that was dilated by a scale factor 3 about the origin. This may help to understand that all linear values after dilation become multiplied by scale factor.

5 0

3 years ago

Read 2 more answers

How to solve by the square root property

nata0808 [166]

First, you want to divide both sides by 3 to isolate your variable. You get x^2 = 9. The square root of x^2 and 9 equals x = 3.

x = 3 is your answer.

4 0

3 years ago

A triangle has side lengths of 9 in, 13 in, and 20 in. What is the measurement of this triangle’s largest angle?

zvonat [6]

1. Given any triangle ABC with sides BC=a, AC=b and AB=c, the following are true :

i) the larger the angle, the larger the side in front of it, and the other way around as well. (Sine Law) Let a=20 in, then the largest angle is angle A.

ii) Given the measures of the sides of a triangle. Then the cosines of any of the angles can be found by the following formula:

$a^{2}=b ^{2}+c ^{2}-2bc(cosA)$

2.

$20^{2}=9 ^{2}+13 ^{2}-2*9*13(cosA)

400=81+169-234(cosA) 150=-234(cosA)

cosA=150/-234= -0.641$

3. m(A) = Arccos(-0.641)≈130°,

4. Remark: We calculate Arccos with a scientific calculator or computer software unless it is one of the well known values, ex Arccos(0.5)=60°, Arccos(-0.5)=120° etc

4 0

4 years ago

Read 2 more answers

URGENT PLS HELP ASAP ANSWER EACH EMPTY BOX AND I WILL GIVE BRAINLIEST !!!!!

Aleks [24]

Answer:

y-1 = 2(x+2)

Step-by-step explanation:

Point slope form of a line is

y-y1 = m(x-x1) where m is the slope and (x1,y1) is a point on the line

y -1 = 2(x - -2)

y-1 = 2(x+2)