All categories

3 years ago

Can someone please take a look at the attachment and tell me what the answer is? Serious answers only please!

Mathematics

1 answer:

siniylev [52]3 years ago

5 0

The solid line indicates that the line itself is included in the solutions. That the shaded area is to the left, or under, the line means that y≤the line.

To find the equation of the line you first find the slope or m using any two points.

m=(y2-y1)/(x2-x1) in this case you are given (2,-5) and (-2, 3) so

m=(3--5)/(-2-2)

m=8/-4

m=-2

So far for y≤mx+b we have:

y=-2x+b using either point we can solve for b, I'll use (2,-5)

-5=-2(2)+b

-5=-4+b

-1=b

So the line is:

y=-2x-1 and since y≤ the line:

y≤-2x-1

You might be interested in

Simplify the expression completely 125 1/3

Roman55 [17]

<h3>I'll teach you how to solve 125 1/3</h3>

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

125 1/3

Convert mixed numbers to improper fractions:

125 1/3=

125*3+1/3=

376/3

Your Answer Is 376/3

plz mark me as brainliest :)

3 0

3 years ago

Suppose the commute times for employees of a large company follow a

Kazeer [188]

The best answer for this question is B

5 0

3 years ago

Read 2 more answers

Find the area of the figure below.<br> A. 288.62<br> B. 266.22<br> C. 186.66<br> D. 236.6

Setler79 [48]

Answer:

Step-by-step explanation:

hope this helps

7 0

3 years ago

If b= 18^3 in the right triangle below, what is the value of c?

dolphi86 [110]

It would be 55 at the most

5 0

2 years ago

Can SOMEBODYYY please do the maths!!! PLeassssssssssssseeeeeeee

Anna11 [10]

Answer:

<h2>see below</h2>

Step-by-step explanation:

<h3>Question-6:</h3>

we are given a equation

$\sf \displaystyle \: \log_{4}( - x) + \log_{4}( 6 - x) = 2$

to solve so

recall logarithm multiplication law:

$\sf \displaystyle \: \log_{4}( - x \times (6 - x)) = 2$

simplify multiplication:

$\sf \displaystyle \: \log_{4}( - 6 x + {x}^{2} ) = 2$

remember $\displaystyle \log_{4}(4^2)=2$

$\sf \displaystyle \: \log_{4}( - 6 x + {x}^{2} ) = \log_{4}( {4}^{2} )$

cancel out $\log_4$ from both sides:

$\sf \displaystyle \: - 6 x + {x}^{2} = {4}^{2}$

simplify squares:

$\sf \displaystyle \: - 6 x + {x}^{2} = 16$

move left hand side expression to right hand side and change its sign:

since we are moving left hand side expression to right hand side there'll be only 0 left in the left hand side

$\sf \displaystyle \: - 6 x + {x}^{2} - 16 = 0$

rewrite it to standard form i.e ax²+bx+c=0

$\sf \displaystyle \: {x}^{2} - 6x - 16 = 0$

rewrite -6x as 2x-8x:

$\sf \displaystyle \: {x}^{2} + 2x - 8x - 16 = 0$

factor out x and 8:

$\sf \displaystyle \: x {(x}^{} + 2) - 8(x + 2) = 0$

group:

$\sf \displaystyle \: (x - 8){(x}^{} + 2) = 0$

$\displaystyle \: x = 8 \\ x = - 2$

$\therefore \: x = - 2$

<h3>Question-7:</h3>

move left hand side log to right hand side:

$\displaystyle \: \log(x ) + \log(x - 21) = 2$

use mutilation logarithm rule;

$\displaystyle \: \log( {x}^{2} - 21x) = 2$

$\log(10^2)=2$ so

$\displaystyle \: \log( {x}^{2} - 21x) = \log({10}^{2} )$

cancel out log from both sides:

$\displaystyle \: {x}^{2} - 21x = 100$

make it standard form:

$\displaystyle \: {x}^{2} - 21x - 100= 0$

factor:

$\displaystyle \: {(x} + 4)(x - 25)= 0$

$\displaystyle \: x = 25$

5 0

2 years ago