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3 years ago

Hello, please help me with this question, thank you very much.

Mathematics

1 answer:

LUCKY_DIMON [66]3 years ago

7 0

Answer:

(A) <BAC = <ABC

(B) <C = 68°

( 2 * <A + <C = 180°)

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In parallelogram ABCD, point M is the midpoint of the side BC , point O is the intersection point of the segment AM and diagonal

Greeley [361]

Answer: BO:OD = 1:2

Step-by-step explanation:

Here, ABCD is parallelogram, point M is the midpoint of the side BC , point O is the intersection point of the segment AM and diagonal BD.

Let N is the mid point AB ⇒ AN=NB or 2.BN=DC. ( because AB=DC)

And join Points N and C. (by construction)

Then, In triangles NOB and DOC,

∠NOB=∠DOC ( vertically opposite angles)

And, ∠OBN = ∠ODC ( By the property of interior alternative angles by same transversal)

Thus, By the property of similarity,

$\triangle NOB\sim \triangle DOC$

Therefore, $\frac{OB}{OD} = \frac{BN}{DC}$ = $\frac{BN}{2BN} = \frac{1}{2}$

5 0

4 years ago

A rule for creating a pattern is given below.

Harrizon [31]

Answer:

(Choice B)

Step-by-step explanation:

(7, 0)

7 - 7 = 0

Here, we can see that when 7 is subtracted from x, we get 0 (our y)

Hope this helps!

8 0

3 years ago

Can someone help with explanation?

ch4aika [34]

Answer:

Step-by-step explanation:

multiply 7 times 4 then you add that when you

add 4+1=5 then you multiply those two

8 0

3 years ago

Help me answer this question please help it is hard

kati45 [8]

Add all of the numbers together , and then divide by how many numbers your adding up. in this case you would divide by 8.

5 0

4 years ago

Solve for 2cos^2theta + 13costheta = -6. theta is >= 0 and < 360 degrees

LiRa [457]

$\\ 2cos^2 \theta+13cos \theta=-6 \\ 2cos^2 \theta+13cos \theta+6=0 \\ 2cos^2 \theta+12cos \theta+1cos \theta+6=0 \\ 2cos \theta(cos \theta+6)+1(cos \theta+6)=0 \\ (2cos\theta +1)(cos\theta+6)=0 \\$

$cos\theta+6=0 \\ cos\theta=-6 \\ \\ cos(x)=|a| \leq 1 \\ |-6| \geq 1 \\ \\$

$2cos \theta+1=0 \\ 2cos \theta=-1 \\ cos \theta=- \frac{1}{2} \\ \theta= \frac{ 2\pi }{3} +2 \pi k~~~k \in Z \\ \\ 0 \leq \theta \ \textless \ 360^{а} \\ 0 \leq \theta \ \textless \ 2 \pi \\ \theta = \frac{2 \pi }{3} \\ \theta = \frac{2 \pi }{3} , \frac{4 \pi }{3}$

7 0

3 years ago