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2 years ago

Angle ABF is 77

Mathematics

1 answer:

Nonamiya [84]2 years ago

4 0

Answer:

a) 77 b)B

Step-by-step explanation:

letterF is the same as 77 meaning that X will be the same as letterF cause the are vertically opposite angles.

I just did the question on math watch.

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6²+8²=12² is this triangle a right triangle? Explain

leonid [27]

No because 36+64≠144

8 0

3 years ago

Place grouping symbols to make this equation true. <br><br> 9 + 2 x 6 - 8 - 2 x 1/2 = 18

saw5 [17]

Answer:

no the answer is 28

Step-by-step explanation:

9+2=11x6=66-8=58-2=56x12= 28 or 56x0.5= 28

6 0

3 years ago

Find the missing y-coordinate that makes the two triangles congruent. Triangle ABC: A(8,4), B(2,6), C(5, 0) Triangle MNO: M(7,4)

Zolol [24]

Answer:

y = 8

Step-by-step explanation:

Two triangles are said to be congruent if all the three sides and three angles of both triangles are equal.

The distance between two points on the coordinate plane is given as:

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\$

In triangle ABC:

$|AB|=\sqrt{(2-8)^2+(6-4^2)}=\sqrt{40} =2\sqrt{10}\\\\|AC|=\sqrt{(5-8)^2+(0-4)^2}=5\\\\|BC|=\sqrt{(5-2)^2+(0-6)^2 }=\sqrt{45}$

In triangle MNO:

$|MN|=\sqrt{(1-7)^2+(2-4^2)}=\sqrt{40} =2\sqrt{10}\\\\|MO|=\sqrt{(4-7)^2+(y-4)^2}\\\\|NO|=\sqrt{(4-1)^2+(y-2)^2 }$

Since triangle ABC and triangle MNO are congruent, hence:

|AB| = |MN| = 2√10, |AC| = |MO| = 5, |BC| = |NO| = √45

$|AC|=|MO|=5\\\\\sqrt{(4-7)^2+(y-4)^2}=5\\\\(4-7)^2+(y-4)^2=25\\\\9 +(y-4)^2=25\\\\(y-4)^2=16\\\\square\ root\ of\ both\ sides:\\\\y-4=4\\\\y=4+4\\\\y=8$

Hence O = (4, 8)

3 0

3 years ago

Can I get some help ? :/

Nastasia [14]

So divide 20 by 3 because there are 3 fractions and you need to get 20 =18
20-18 =2=1/1/1x2 =your anwser

3 0

3 years ago

23.

Crazy boy [7]

Answer:

The number that cannot be the largest possible 6-digit number is;

(D) AAABCB

Step-by-step explanation:

From the question, we have;

A, B, and C = Distinct digits, therefore, A ≠ B ≠ C

The number of digits in the number to be formed = 6 digits

The number of 'A' in the number to be formed = 3

The number of 'B' in the number to be formed = 2

The number of 'C' in the number to be formed = 1

We have;

When A > B > C

The largest possible number = AAABBC

When C > A > B

The largest possible number = CAAABB

When B > A > C

The largest possible number = BBAAAC

When A > C > B

The largest possible number = AAACBB

Therefore, given that when A > B > C, the largest possible number = AAABBC, we have;

AAABBC > AAABCB, because B > C, therefore, within the tens and unit of the two 6 digit numbers, we have, BC > CB

∴ AAABBC > AAABCB and <u>AAABCB</u>, cannot be the largest possible 6-digit number

3 0

3 years ago