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3 years ago

Pls Help!!!!! Directions: Find all missing angles.

Mathematics

1 answer:

garik1379 [7]3 years ago

8 0

Answer:

m<1 = 87°

m<2 = 45°

m<3 = 45°

m<4 = 52°

Step-by-step explanation:

Angle C is a right angle. Therefore, it is 90°. Because line segment BC bisects angle C, the two angles would be 45° since 90 ÷ 2 = 45.

So, angles 2 and 3 equal 45°.

Now that we have two angle measures, we can find the measure of the third angle.

45 + 48 = 93°

180 - 93 = 87°

Angle 1 equals 87°.

45 + 83 = 128°

180 - 128 = 52°

Angle 4 equals 52°.

Hope that helps.

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Which of the following slopes of a line passes through points (2, 3) and (4,7) ?

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Answer:

Step-by-step explanation:

slope = (difference in y)/(difference in x)

slope = (7 - 3)/(4 - 2)

slope = 4/2 = 2

Answer: 2

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3 years ago

The velocity v that an object r units

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Answer:

M=v^2r / 2G

Step-by-step explanation:

Open the file, it is solved for you, your welcome.

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3 years ago

Slope and y-intercept form from a table

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Answer:

The y-intercept is 3

Gradient: 4

11-3= 8

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2 years ago

The first three terms of a sequence are given. Round to the nearest thousandth (if necessary).

BabaBlast [244]

Step-by-step explanation:

the introduction of a fraction tells us that we are dealing with multiplications, and therefore a geometric sequence (where every new term is created by multiplying the previous term by a constant factor, the ratio r).

I think your teacher made a mistake, or you made one when typing the question in here.

there is no factor r that creates

15×r = 9

and

9×r = 5/27

it would mean that

15 × r² = 5/27

r² = 5/27 / 15 = 5/27 × 1/15 = 5/405 = 1/81

r = 1/9

but 15 × 1/9 = 5 × 1/3 = 5/3 is NOT 9

and 9 × 1/9 = 9/9 = 1 is NOT 5/27

so, this can't be right.

on the other hand

15 × r = 9

r = 9/15 = 3/5

and then

9 × 3/5 = 27/5

so, either the sequence should have been

15, 5/3, 5/27

or (and I suspect this to be true)

15, 9, 27/5

under that assumption we have

s1 = 15

r = 3/5

sn = sn-1 × r = s1 × r^(n-1) = 15 × (3/5)^(n-1)

s10 = 15 × (3/5)⁹ = 15 × 19683/1953125 =

= 3 × 19683/390625 = 59049/390625 =

= 0.15116544 ≈ 0.151

6 0

2 years ago

The average number of traffic tickets issued in a city on any given day

Anton [14]

Answer:

The most tickets were written on Saturday .On Saturday 325 tickets were issued

Step-by-step explanation:

The average number of traffic tickets issued in a city on any given day Sunday-Saturday can be approximated by

$T(x) = -6x^2 + 84x + 37$

Where x represents the number of days after Sunday

T(x) represents the number of traffic tickets issued.

Sunday = x=0

Monday = x=1

Tuesday = x=2

Wednesday = x=3

Thursday = x =4

Friday = x=5

Saturday = x=6

Substitute x= 0

$T(x) = -6x^2 + 84x + 37\\T(x) = -6(0)^2 + 84(0) + 37\\T(x)=37$

On Sunday 37 tickets were issued

Substitute x= 1

$T(x) = -6x^2 + 84x + 37\\T(x) = -6(1)^2 + 84(1) + 37\\T(x)=115$

On Monday 115 tickets were issued

Substitute x= 2

$T(x) = -6x^2 + 84x + 37\\T(x) = -6(2)^2 + 84(2) + 37\\T(x)=181$

On Tuesday 181 tickets were issued

Substitute x= 3

$T(x) = -6x^2 + 84x + 37\\T(x) = -6(3)^2 + 84(3) + 37\\T(x)=235$

On Wednesday 235 tickets were issued

Substitute x= 4

$T(x) = -6x^2 + 84x + 37\\T(x) = -6(4)^2 + 84(4) + 37\\T(x)=277$

On Thursday 277 tickets were issued

Substitute x= 5

$T(x) = -6x^2 + 84x + 37\\T(x) = -6(5)^2 + 84(5) + 37\\T(x)=307$

On Friday 307 tickets were issued

Substitute x= 6

$T(x) = -6x^2 + 84x + 37\\T(x) = -6(6)^2 + 84(6) + 37\\T(x)=325$

On Saturday 325 tickets were issued

Hence the most tickets were written on Saturday .On Saturday 325 tickets were issued

6 0

3 years ago