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3 years ago

In triangle ABC the length of AC is 15cm.

Mathematics

1 answer:

ZanzabumX [31]3 years ago

6 0

Answer:

35°

Step-by-step explanation:

Total of all angles is 180°

Angles A+B+C=180°

33°+112°+C=180°

180°-145°=Angle C

C=35°

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Use exponents to write 81 in 3 different ways

denis23 [38]

9 to the second power 3 to the fourth power and 81 to the first power

8 0

3 years ago

Rory earned an 84% on his test. He answered 21 questions correctly. How many total questions were on the test? Round to the near

Free_Kalibri [48]

Answer:

25 questions

Step-by-step explanation:

84%*x = 21

-simplify-

0.84 *x = 21

Divide both sides by 0.84 than...

you get x = 25

6 0

3 years ago

(60 POINTS) Please explain this in detail

Alik [6]

H=D

O=C

B=M

R=A

M=B

First we will prove that the diagonals are bisectors of each other.

Angle DBA is congruent to angle BDC.

Angle CMD is congruent to angle AMB.

Triangle CMD is congruent to triangle AMB.

Segment AM is congruent to segment MC.

M is the midpoint of segment AC.

Segment BD bisects segment AC.

Segment BM is congruent to segment MD.

M is the midpoint of segment BD.

Segment AC bisects segment BD.

Then we will prove that the diagonals are perpendicular to each other.

10. Segment CD is congruent to segment CB.

11. Segment BM is congruent to segment MB.

12. Segment CM is congruent to segment CM.

13. Triangle CDM is congruent to triangle CBM.

14. Angle CMD is congruent to triangle BMC.

15. Angle CMD + angle BMC = 180

16. Angle CMD = 90 and angle BMC = 90

17. Angle AMB = 90

18. Angle CMD is congruent to angle DMA.

19. Angle DMA = 90

20. Segment AC is perpendicular to segment BD.

5 0

3 years ago

Read 2 more answers

I need help with 12 13 and 14

nikdorinn [45]

Answer: Lines $\frac{}{BC}$ and $\frac{}{EF}$ are different lengths.

Step-by-step explanation:

The distance formula is $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }$ , and you can use this formula to solve for the lengths of both lines $\frac{}{BC}$ and $\frac{}{EF}$ .

For line $\frac{}{BC}$ , let $x_{1}$ = the x at point B, or 1, and let $x_{2}$ = the x at point C, or 2.

Now, let $y_{1}$ = the y at point B, or 4, and let $y_{2}$ = the y at point C, or -1.

Now, solve the formula to find the length $\frac{}{BC}$ = $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\$ .

$\frac{}{BC}$ = $\sqrt{(1-2)^{2} +(4-(-1))^2$

$\frac{}{BC}$ = $\sqrt{(-1)^{2} +(4+1)^2$

$\frac{}{BC}$ = $\sqrt{1 +5^2$

$\frac{}{BC}$ = $\sqrt{(1+25)$

$\frac{}{BC}$ = $\sqrt{26} \\$

Now, for line $\frac{}{EF}$ , let $x_{1}$ = the x at point E, or -4, and let $x_{2}$ = the x at point F, or -1.

Let $y_{1}$ = the y at point E, or -3, and let $y_{2}$ = the y at point F, or 1.

Now, solve the formula to find the length $\frac{}{EF}$ = $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\$ .

$\frac{}{EF}$ = $\sqrt{(-4-(-1))^{2} +(-3-1)^2$

$\frac{}{EF}$ = $\sqrt{(-4+1)^{2} +(-4)^2$

$\frac{}{EF}$ = $\sqrt{(-3)^2+16$

$\frac{}{EF}$ = $\sqrt{(9+16)$

$\frac{}{EF}$ = $\sqrt{25}$

$\frac{}{EF}$ = 5

Now, look back at $\frac{}{BC}$ . The two lines have different lengths, so you have now justified the fact that they are not the same.

Questions 13 and 14 would be solved in much the same way- but please let me know if you want me to show the work for those as well!

7 0

3 years ago

Factor w^4x^2y^3−w^2x^2y WITHOUT using grouping

gavmur [86]

Answer:

w^2x^2y * (wy-1)*(wy+1)

Step-by-step explanation:

hope this helps!

3 0

3 years ago