All categories

3 years ago

I'm not sure how to do this! 50 points?

Mathematics

2 answers:

Rus_ich [418]3 years ago

6 0

Answer:

MOP=33

Step-by-step explanation:

MOP + MON = NOP

Substitute what we know

MOP = 3x-3

MON = 4x+9

NOP = 90 right angle

3x-3+4x+9 = 90

Combine like terms

7x+6 =90

Subtract 6 from each side

7x+6-6=90-6

7x = 84

Divide each side by 7

7x/7 = 84/7

x =12

We need to find MOP

MOP = 3x-3

=3(12)-3

36-3

galben [10]3 years ago

4 0

Answer:

MOP=33

Step-by-step explanation:

MOP + MON = NOP

Substitute what we know

MOP = 3x-3

MON = 4x+9

NOP = 90 right angle

3x-3+4x+9 = 90

Combine like terms

7x+6 =90

Subtract 6 from each side

7x+6-6=90-6

7x = 84

Divide each side by 7

7x/7 = 84/7

x =12

We need to find MOP

MOP = 3x-3

=3(12)-3

36-3

---------------------

I hope this is correct for u

You might be interested in

3 3/4 ÷ (1 2/3 + 2 1/2)

ZanzabumX [31]

Answer:

9/10.

Step-by-step explanation:

3 3/4 ÷ (1 2/3 + 2 1/2)

= 3 3/4 / ( 5/3 + 5/2)

= 3 3/4 / ( 10/6 + 15/6)

= 3 3/4 / 25/6

= 15/ 4 / 25/6

= 15/4 * 6/25

= 90/100

= 9/10.

4 0

2 years ago

if two angles of a triangle have measures of 42 degrees and 20 degrees what is the measure of the third

Mila [183]

Every triangle has an angle measure of 180 degrees.

so 42+20+x (where x is the measure of the third angle)=180

Combine like terms.

62+x=180

Subtract 62 from both sides

x=118.

3 0

3 years ago

What does 3 over 4 times 2 equal?

Hoochie [10]

Answer: 1.5

Da maths: Bruh ask google

6 0

3 years ago

Read 2 more answers

Consider the sequence defined recursively by do = 0, an = an-1 + 3n – 1. a) Write out the first 5 terms of this sequence,

creativ13 [48]

Answer:

The first 5 terms of the sequence is 2,7,15,26,40.

Step-by-step explanation:

Given : Consider the sequence defined recursively by $a_0=0$ $a_n=a_{n-1}+3n-1$

To find : Write out the first 5 terms of this sequence ?

Solution :

$a_n=a_{n-1}+3n-1$ and $a_0=0$

The first five terms in the sequence is at n=1,2,3,4,5

For n=1,

$a_1=a_{1-1}+3(1)-1$

$a_1=a_{0}+3-1$

$a_1=0+2$

$a_1=2$

For n=2,

$a_2=a_{2-1}+3(2)-1$

$a_2=a_{1}+6-1$

$a_2=2+5$

$a_2=7$

For n=3,

$a_3=a_{3-1}+3(3)-1$

$a_3=a_{2}+9-1$

$a_3=7+8$

$a_3=15$

For n=4,

$a_4=a_{4-1}+3(4)-1$

$a_4=a_{3}+12-1$

$a_4=15+11$

$a_4=26$

For n=5,

$a_5=a_{5-1}+3(5)-1$

$a_5=a_{4}+15-1$

$a_5=26+14$

$a_5=40$

The first 5 terms of the sequence is 2,7,15,26,40.

5 0

3 years ago

Which numbers are 4 units from -2 on the number line

Artemon [7]

Let the number be x

Then the absolute value of the number from -2 should be 4

$|x--2|=4$

$|x+2|=4$

So either

$(x+2)=4$

$x=4-2=2$

$-(x+2)=4$

$x+2=-4$

$x=-4-2$

$x=-6$

[tex] <B>The numbers are 2 and -6</B> [/tex[

They are four units away from -2 on the number line

4 0

3 years ago