All categories

2 years ago

HELP ASAP IF YOU TAKE THE POINTS IM REPORTING U QUESTION IN PIC

Mathematics

2 answers:

Andrew [12]2 years ago

7 0

Let's call the number x. So four times the number minus five would look like
4x-5
Twice the number plus three would look like
2x+3
Now make the two equations equal
4x-5=2x+3
I'm going to subtract 2x from each side
2x-5=3
Now I'm going to add five to each side
2x=8
Now divide both sides by 2
x=4
To double check, just plug back into the original equations
4(4)-5=11
2(4)+3=11
They are equal, so the answer is x=4.

Airida [17]2 years ago

4 0

Step-by-step explanation:

Let the number be x,

Then, according to the question,

4x-5=2x+3

(we need to find the value of x)

4x-2x=5+3

= 2x=8

= x=8/2

= x=4

You might be interested in

One angle of an isosceles triangle measures 118°. Which other angles could be in that

jonny [76]

Answer:

that is the solution to the question

8 0

2 years ago

A tree struck by lighting broke at a point 16 feet above ground level. The broken trunck formed a right triangle with the top of

pantera1 [17]

Answer:

okay i need to know the options

Step-by-step explanation:

4 0

3 years ago

Lori is three times as old as Michelle. In five years Michelle's age will be one half of Lori's age. How old are Lori and Michel

AURORKA [14]

Answer:

Lori's age is 15 years old and Michelle's age is 5 years old now

Step-by-step explanation:

Let

x-----> Lori's age

y ----> Michelle's age

we know that

$x=3y$ -----> equation A

$(y+5)=\frac{1}{2}(x+5)$ ----> equation B

substitute equation A in equation B and solve for y

$(y+5)=\frac{1}{2}(3y+5)$

$2y+10=(3y+5)$

$3y-2y=10-5$

$y=5$

Find the value of x

$x=3(5)=15$

therefore

Lori's age is 15 years old and Michelle's age is 5 years old now

7 0

3 years ago

The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?

Lina20 [59]

$\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
\\
\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0$
$\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y

$
$y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
\\
\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
\\
\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
\\
\\y' = \frac{3}{( -32 )}$

7 0

3 years ago

Al Bank is worried about financing of his new home. The house sells for $490,000. If he puts down 20 percent, what will Al's mon

Anastaziya [24]

The answer is A,
how much interest:3675

4 0

3 years ago