MATHEMATICS HIGH SCHOOL

Help please ahahah i hate this sm i really need help

Answers

Answer 1

Answer:

Answer:

-7/6

Step-by-step explanation:

- take two values of x and their corresponding two values of y

- use the slope formula ( y2 - y1 )/( x2 - x1 )

- i used x values 4 and 10 but you can use any

( -6 - (-13) ) / ( 4 - 10 ) = -7/6

Answer 2

Answer: It should be -7/6 hope you got it correct

What is equation?

An equation is arrangement of variables and coefficient which provides to understand given statement in form of numbers .That means a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value.

How to solve?

Given equation y=5x+10 and other equation y=5x+2

to check if two have any solution possible we will substitute value and check if RHS=LHS or not

∴y=5x+10=5x+2

=5x+10=5x+2

⇒0= -8

hence RHS≠ LHS hence no solution is possible

Learn more about finding solutions brainly.com/question/1365672

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The system of equations
$\left \{ {{y = 5x + 10 } \atop { y = 5x + 2}} \right.$

you can easily see that they are contrary to the equation . We not solve them . However prove it.

by substitution .
We substitute the value of y from the first equation to the second equation

y = 5x + 10
y = 5x + 2

y = 5x + 10
5x + 10 = 5x + 2

y = 5x + 10
5x - 5x = 2 - 10

y = 5x + 10
0 ≠ -8

or by opposing coefficients
y = 5x + 10 [* (-1)
y = 5x + 2

-y = -5x - 10
y = 5x + 2
----------------- sum
0 ≠ -8

Answer. No solution

Find the distance between the two ponts rounding to the nearest tenth (if necessary)(3,6) and (1,9)

Answers

i think it is 3.6
hope this helps

What is the solution to the equation 1/2 (x-12) + 4 = 1/3 (6x-15)?

Answers

Answer:

$x=2$

Step-by-step explanation:

$(1)/(2) (x-12) + 4 = (1)/(3) (6x-15)$

$(x-12)/(y2) +4=(6x-15)/(3)$

$((x-12)+2*4)/(2)=((2*3)x-3*5)/(3)$

$x=2$

Log4(x)+log4(x+15)=2

Answers

$D:x>0 \wedge x+15>0\nD:x>0 \wedge x>-15\nD:x>0\n\n\log_4x+\log_4(x+15)=2\n\log_4x(x+15)=2\n4^2=x(x+15)\n16=x^2+15x\nx^2+15x-16=0\nx^2-x+16x-16=0\nx(x-1)+16(x-1)=0\n(x+16)(x-1)=0\nx=-16 \vee x=1\n-16\not \in D \Rightarrow \boxed{x=1}$