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3 years ago

Find the value of x.

Mathematics

2 answers:

Arlecino [84]3 years ago

5 0

Answer:

x=80

Step-by-step explanation:

a circle is 360 degrees and 125+155 is 280 and 360-280=80

marta [7]3 years ago

3 0

Answer:

x = 80 degrees

Step-by-step explanation:

To answer this question, we need to first remember that a full circle equals 360 degrees, so therefore, in order to answer this question all we have to do is some addition and some subtraction.

First, we can add 125 and 155 together to get:

125 + 155 = 280

Then, we subtract 360 by 280:

360 - 280 = 80

And now we have our answer, which is x = 80 degrees

Hope this helps :)

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a pile of sand has a weight of 90kg The sand is put into a small bag, a medium bag and a large bag in the ratio of 2 : 3 : 7 Wor

just olya [345]

Hi there!

To split 90 kilos in the ratio of 2 : 3 : 7 we must first realise that we have a total of 2 + 3 + 7 = 12 parts, in which we must split the total 90 kilos.

12 parts equal 90 kilo, and therefore
1 part equals 90 / 12 = 7.5 kilos.

1 part equals 90 / 12 = 7.5 kilos, and therefore
2 parts equal 7.5 × 2 = 15 kilos.

1 part equals 90 / 12 = 7.5 kilos, and therefore
3 parts equal 7.5 × 3 = 22.5 kilos.

1 part equals 90 / 12 = 7.5 kilos, and therefore
7 parts equal 7.5 × 7 = 52.5 kilos.

Hence, 90 kilos in the ratio of 2 : 3 : 7
gives 15 kg, 22.5 kg and 52.5 kg.

~ Hope this helps you!

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3 years ago

3. In an auditorium, 1/6 of the students are fifth graders, 1/3 are fourth graders, and 1/4 of the remaining students are second

Katen [24]

<h3>Further explanation</h3>

<u>Given:</u>

In an auditorium,

$\frac{1}{6}$ of the students are fifth graders,
$\frac{1}{3}$ are fourth graders, and
$\frac{1}{4}$ of the remaining students are second graders.

There are 96 students in the auditorium.

<u>Question:</u>

How many second graders are there?

<u>The Process:</u>

The least common multiple (LCM) of 3 and 6 is 6.

Let us draw the diagram of all students.

$96 \div 6 \ units = 16 \rightarrow \boxed{16}\boxed{16}\boxed{16}\boxed{16}\boxed{16}\boxed{16} = 96 \ students$

$\frac{1}{6}$ of the students are fifth graders, or 1 of 6 units above, that is $\boxed{16} = 16 \ students$
$\frac{1}{3} = \frac{2}{6}$ are fourth graders, or 2 of 6 units above, that is $\boxed{16}\boxed{16} = 32 \ students$

The remainder is 6 units - (1 unit + 2 units) = 3 units, that is $\boxed{16}\boxed{16}\boxed{16} = 48 \ students$ . Or, 96 - 16 - 32 = 48 students.

$\frac{1}{4}$ of the remaining students are second graders.

Let us count how many second graders are there.

$\boxed{ \ \frac{1}{4} \times 48 \ students = ? \ }$

Thus, there are 12 second-graders in the auditorium.

- - - - - - - - - -

Quick Steps

$\boxed{ \ \frac{1}{4} \times \bigg(1 - \frac{1}{6} - \frac{1}{3} \bigg) \times 96 = ? \ }$

$\boxed{ \ = \frac{1}{4} \times \bigg(\frac{6}{6} - \frac{1}{6} - \frac{2}{6} \bigg) \times 96 \ }$

$\boxed{ \ = \frac{1}{4} \times \frac{3}{6} \times 96 \ }$

$\boxed{ \ = \frac{3}{24} \times 96 \ }$

We crossed out 24 and 96.

$\boxed{ \ = 3 \times 4 \ }$

$\boxed{\boxed{ \ = 12 \ students \ }}$

<h3>Learn more</h3>

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Keywords: In an auditorium, 1/6 of the students, fifth graders, 1/3, fourth, 1/4, the remaining students, second, 96, how many, second graders

3 0

3 years ago

Read 2 more answers

What is the equation, in standard form of the parabola that contains the following points? (-2,18), (0,2), (4,42)

Vladimir79 [104]

Answer: y = 3 $x^{2}$ - 2x + 2

Step-by-step explanation:

The equation in standard form of a parabola is given as :

y = $ax^{2}$ + bx + c

The points given are :

( -2 , 18 ) , ( 0,2) , ( 4 , 42)

This means that :

$x_{1}$ = -2

$x_{2}$ = 0

$x_{3}$ = 4

$y_{1}$ = 18

$y_{2}$ = 2

$y_{3}$ = 42

All we need do is to substitute each of this points into the equation , that is , $x_{1}$ and $y_{1}$ will be substituted to get an equation , $x_{2}$ and $y_{2}$ will be substituted to get an equation and $x_{3}$ , $y_{3}$ will also be substituted to get an equation also.

Starting with the first one , we have :

y = $ax^{2}$ + bx + c

18 = a[ $(-2)^{2}$ ] + b (-2) + c

18 = 4a - 2b + c

Therefore :

4a - 2b + c = 18 ................ equation 1

substituting the second values , we have

2 = a (0) + b ( 0) + c

2 = c

Therefore c = 2 ............... equation 2

also substituting the third values , we have

42 = a[ $(4)^{2}$ ] + b (4) + c

42 = 16a + 4b + c

Therefore

16a + 4b + c = 42 ........... equation 3

Combining the three equations we have:

4a - 2b + c = 18 ................ equation 1

c = 2 ............... equation 2

16a + 4b + c = 42 ........... equation 3

Solving the resulting linear equations:

substitute equation 2 into equation 1 and equation 3 ,

substituting into equation 1 first we have

4a - 2b + 2 = 18

4a - 2b = 16

dividing through by 2 , we have

2a - b = 8 ............... equation 4

substituting c = 2 into equation 3 , we have

16a + 4b + c = 42

16a + 4b + 2 = 42

16a + 4b = 40

dividing through by 4 , we have

4a + b = 10 ................ equation 5

combining equation 4 and 5 , we have

2a - b = 8 ............... equation 4

4a + b = 10 ................ equation 5

Adding the two equations to eliminate b , we have

6a = 18

a = 18/6

a = 3

Substituting a = 3 into equation 4 to find the value of b , we have

2(3) - b = 8

6 - b = 8

b = 6 - 8

b = -2

Therefore :

a = 3 , b = -2 and c = 2

Substituting these values into the equation of parabola in standard form , we have

y = 3 $x^{2}$ - 2x + 2

3 0

4 years ago