Ask question

Ask question

All categories

3 years ago

15

Solve for x and y Hi

1 answer:

irakobra [83]3 years ago

5 0

Answer:

x = 10, y = 20

Step-by-step explanation:

Because two sides of the rightmost triangle are congruent, the base angles would follow suit. This means 5x - 10 = 3x + 10. Simplifying this we get:

2x = 20

x = 10

Next, in the other triangle, we can say the same thing, the base angles would be congruent. From this we can say that the other base angle is 4y. Using the exterior angle theorem, we can say that the two angles with x terms add to 4y. This means 3x + 10 + 5x - 10 is 4y. We already know x is 10, so substituting that in we get:

80 = 4y

y = 20

You might be interested in

Which of the binomials below is a factor of this trinomial?

iren [92.7K]

Answer: i think its x-4

Step-by-step explanation:

i

5 0

3 years ago

Solve the equation 4(3y-1)-3(2y)=6

olga nikolaevna [1]

Answer:

$4(3y - 1) - 3(2y) = 6$

Remove backets

$12y - 4 - 6y = 6$

Group and Evaluate like terms

$12y - 6y = 6 + 4$

$6y = 10$

$y = 10 \div 6$

$y = 1.67$

$y = \frac{5}{3}$

6 0

4 years ago

The area of the trapezoid is 39 ft2. What is the height of the trapezoid?

xeze [42]

A= 1/2 h (b1+b2)
84=1/2h (9+15)=12h
h=7ft

8 0

3 years ago

Trigonometry is the study of tree-dimensional objects true or false

Fofino [41]

False, trigonometry is not the study of three dimensional shapes.

6 0

3 years ago

AREA ADDITION AND SUBTRACTION ASSISTANCE?

Tom [10]

The shaded area is a segment. The area of a segment is

$A_{segment}=A_{sector}-A_{triangle}$ .

You know radius that is the side of isosceles triangle and measure of angle between two equal triangle sides, then

$A_{triangle} =\dfrac{1}{2} a\cdot a \sin \alpha=\dfrac{1}{2} 18.6\cdot 18.6 \cdot \sin 123^{\circ}=172.98\cdot \sin 123^{\circ}=145.07$ sq. m..

The area of sector is $A_{sector}=\dfrac{r^2\cdot \alpha}{2}= \dfrac{(18.6)^2\cdot 0.68\pi}{2}=117.63\pi=369.53$ sq. m..

Then $A_{segment}=369.53-145.07=224.46$ sq. m.

Answer: $A_{segment}=224.46$ sq. m.

7 0

4 years ago