All categories

4 years ago

Need help with as many possible

Mathematics

1 answer:

snow_lady [41]4 years ago

6 0

M<2 = 89
m<7 = m<2 = 89
m<3 = 180 - 89 = 91
m<8 = m<3 = 91
m<10 =23
m<11 = 62
m<9 = 180 - (62 + 91)
m<9 = 27
m<14 = 1/2 (63+18)
m<14 = 40.5
m<13 = 180 -40.5 = 139.5
m<12 = 180 - (57 + 63) = 60
m<17 = 180 - (60 + 40.5) = 79.5
m<27 = 90
m<25 = 63/2 = 31.5
m<24 = 57/2 = 28.5
m<26 = 90 - (31.5 + 28.5) = 30
m<19 = 180 - (30 + 79.5) = 70.5
m<18 = 180 - 70.5 = 108.5
m<15 = (60+23+63)/ 2 = 73
m<20 = 60/2 =30
m<21 = 90 - 30 = 60
m<22 = 180 -(70.5 + 60) = 49.5
m<23 = 180 -(31.5 + 108.5) = 40
62 + 23 + 63 + 57 = 205
60 + 18 = 78
m<16 = 1/2(205 - 78) = 63.5
m<4 = 1/2(62-13) = 24.5
m<5 = 180 - 24.5 = 155.5
m<6 = m<5 = 155.5
m<1 = 180 -(24.5 + 91) = 64.5

You might be interested in

Plez halp with question bc im smol brain

Studentka2010 [4]

Answer:

B. y= 2x–15

Step-by-step explanation:

The equation that represents the line that passes through the points (8, 1) and (10,5) is y= 2x–15

"x" is found on the left hand side of the bracket and "y" is found on the right hand side of the bracket :)

(8, 1) and (10, 5)

x in the first bracket is 8 and x in the second bracket is 10.

y in the first bracket is 1 and y in the second bracket is 5.

When this equation is applied to find the value for y we get:

y= 2x-15

y= 2(8)-15

y= 16-15

y= 1

y= 2x-15

y= 2(10)-15

y= 20-15

y= 5

We got back the values of y and this means that the equation was correctly chosen and the answer is correct :)

6 0

3 years ago

Brayden, Howard, and Vincent are on the football team. During a game, Brayden gains 1 less than 5 times the average number of ya

Gnoma [55]

Answer: If two pair is parallel then it has no solution.

Step-by-step explanation:

Since we have given that

Let the number of gained yards by each player be y

For the case of Brayden:

Equation will be

$y=5x-1\\\\y-5x=-1$

For the case of Howard :

Equation will be

$y=5x+1\\\\y-5x=1$

For the case of Vincent :

$y=4x+5\\\\y-4x=5$

Since First two equations are parallel so it has no solution.

Reason:

$y-5x=1\\\\y-5x=-1\\\\\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\\\\frac{1}{1}=\frac{5}{5}\neq \frac{1}{-1}$

Hence, if two pair is parallel then it has no solution.

8 0

3 years ago

Read 2 more answers

Which equation can be used to find 30 percent of 600

agasfer [191]

A/b *100
For example,
30/600.
30/600=0.05
0.05*100=5%

5 0

4 years ago

(x+3)(x-2) <br> Polynomial in standard form

postnew [5]

You FOIL the binomials out to get a quadratic.

7 0

3 years ago

Read 2 more answers

The graph shows the relationship between time

jekas [21]

the slope goes by several names

• average rate of change

• rate of change

• deltaY over deltaX

• Δy over Δx

• rise over run

• gradient

• constant of proportionality

however, is the same cat wearing different costumes.

$(\stackrel{x_1}{4}~,~\stackrel{y_1}{64})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{96}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{96}-\stackrel{y1}{64}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{4}}}\implies \cfrac{\stackrel{bottles}{32}}{\underset{mins}{2}}\implies \cfrac{\stackrel{bottles}{16}}{\underset{mins}{1}}$

3 0

3 years ago

Read 2 more answers