MATHEMATICS MIDDLE SCHOOL

Can someone please help me and teach me how to do this !!!!x+y= 11
-x=-y-9

O A. (10,1)
O B. (11,0)
O C. (9,2)
O D. (8, 3)

Answers

Answer 1

Answer:

Answer:

A. ( 10, 1 )

Step-by-step explanation:

Isolate x for x + y = 11.

x + y = 11

Subtract y from both sides.

x + y - 11 - y

Simplify.

x = 11 - y

Substitute x = 11 - y

[ - ( 11 - y ) = -y - 9 ]

Isolate y for - ( 11 - y ) = -y - 9.

- ( 11 - y ) = -y - 9

Expand - ( 11 - y )

Add 11 to both sides.

-11 + y + 11 = - y - 9 + 11

Simplify.

y = -y + 2

Add y to both sides.

y + y = -y + 2 + y

Simplify.

2y = 2

Divide both sides by 2.

$(2y)/(2) =(2)/(2)$

Simplify.

y = 1

Substitute y = 1.

x = 11 - 1

Subtract the numbers: 11 - 1

= 10

x = 10

y = 1

( 10, 1 )

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(6.2)(0.48) What does this mean and equal?

The top letter is “L”
Please help!! Answer correctly and no guessing, please!

Answers

The first thing you must do is find KL. You use a^2 + b^2 = c^2 to do that.

Then you find the various sines and cosines and Tangents of the angles involved.

Step One

Find KL

a^2 + b^2 = c^2

c = 219

b = 178

a = ??

219^2 = 178^2 + a^2

47061 = 31684 + a^2

a^2 = 48061 - 31684

a^2 = 127.97

Step Two.

Find the sines and cosines of J and L

Choice A

Sin(J) =opposite/hypotenuse = 127/219 = 0.5799

Cos(L) = adjacent/hypotenuse = 127/219 = 0.5799

Conclusion Sin(J) = Cos(L) A is not true. Neither is bigger than the other.

Choice B

Choice B does exactly what Choice A did. The Sin of L is the same thing as the cos(J).

Choice C

C can't be true.

Tan(J) = opposite / Adjacent. = 127/178

Tan(L) = opposite / Adjacent = 178/127

One is larger than 1 and the other one is less than 1. They use the same values but in a different order.

We discussed this for choice A. So D is the correct answer

First you need to find out what K and L's angle is

Step #1: Finding L's angle number

to do this you are going to subtract 219-178=?

answer should be 41.

Step #2: Desiding if J is Larger than L

Is 219 Larger than 41? or is 41 larger than 219?

The circumference of a circular stage is 157 feet. If Chris paints the stage at a rate of 400 square feet per hour, how long would it take to the nearest hour, for Chris to paint the entire stage?

Answers

We need the circumference and area of the circle equations:
c = 2*pi*r
a = pi*r^2
lets find the radius from the first equation:
r = c/2*pi
r = 157/(2*3.14)
r = 24.99 ft
now lets calculate the area:
a = pi*r^2
a = 3.14*(24.99)^2
a = 1960.93 ft^2
Now we need to divide the total area by the rate at which Chris paints, and that gives us the time needed:
time = 1960.93/400
time = 4.9 hours
that would be 5 hours rounded to the nearest hour

Expand and simplify (x+2)(x+1)

Answers

So first you would need to FOIL the problem (first-outer-inner-last) so you would multiply x by x and get {x²} then you would multiply x by 1 so you would get just plain {x} then multiply 2 by x so to get {2x} then multiply the last terms 2 and 1 together to get {2} so then put it all together = x² + x + 2x + 2 then you add like terms together so you end up with = x² + 3x + 2 as the answer

Here, we have to FOIL.

(x+2)(x+1) = x^2+3x+2

HELP ME PLEASE, I'M GIVING YOU ALL MY POINTS ;-; (which is 25)just these 3 questionz!!!!!!!!!

Answers

The answer to 31,096 + 59,721 is 89,817 so B 89,800 is the best estimate

2nd- The answer to 69,896 - 42,901 is 26,995 so A 26,900 is the best answer

3rd- 630,429 - 27,811 is 602,618 so the best answer(s) would be A 630,400 and C 27,800

17 points!!!
What is the missing exponent?

(2^-5) ????? = 2^-15

Answers

$(2^-^5) ^x = 2 ^ - ^1^5 \n \n ( (1)/(32) ) ^x = (1)/(32768) \n \n \hbox{Solve the exponent.} \n \n ( (1)/(32))^x = (1)/(32768) \n \n\hbox{Take log of both sides} \n \n \hbox{log}} ( (1)/(32)) ^ x = log ( (1)/(32768)) \n \n x * log ((1)/(32) ) = log ( (1)/(32768) ) \n \n \bold{ (log( (1)/(32768)))/(log (1)/(32) ) } \n \n x = 3$

Therefore the missing exponent is 3. Which makes the equation true.

${ \left( { 2 }^( -5 ) \right) }^( x )={ 2 }^( -15 )\n \n { 2 }^( -5x )={ 2 }^( -15 )\n \n \therefore \quad -5x=-15\n \n \therefore \quad x=3$

Remember that:

${ \left( { a }^( m ) \right) }^( n )={ a }^( m\cdot n )$

What is the more common name for an angle measurer