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faust18 [17]
3 years ago
12

How do I solve 8x+3(x+5)-5(x-4)

Mathematics
1 answer:
adell [148]3 years ago
7 0

Answer:

6x + 35

Step-by-step explanation:

First Distribute

8x + 3x + 15 -5x +20

Then collect like terms

6x + 35

to solve you would need the value of x to plug it!

You might be interested in
Find the value of x in the isosceles triangle shown.​
strojnjashka [21]

Answer:

x = 7

Step-by-step explanation:

Given the base of the triangle is 6 and the line dividing the triangle is x. To get x divide the triangle into two equal halves. When you do so,you’ll have a right triangle with x as the opposite , 8 as the hypotenuse and the adjacent as 3.

Using Pythagoras theorem, we have

8^2 = x^2 + 3^2

8 x 8 = x^2 + 3 x 3

64 = x^2 + 9

Subtract 9 from both sides

64 - 9 = x^2 + 9 - 9

55 = x^2

X^2 = 55

x = squared root of 55

x = 7.4

x = 7

8 0
3 years ago
Evaluate if x =3 and y = -2<br><br> x2+2xy−4y
Marysya12 [62]

Answer:

5

Step-by-step explanation:

With the equation x^2 + 2xy - 4y, we can substitute the values of x = 3 and y = -2 into the equation to find its result.

(3^2) + (2\cdot3\cdot-2) - (4\cdot-2)\\\\9 + (6\cdot-2) - (-8)\\9 - 12 + 8\\5

Hope this helped!

8 0
3 years ago
Help me ill mark u brainliest xd
AlexFokin [52]
20 x 42= 840
So the area would be 840 mm^2.
6 0
2 years ago
Read 2 more answers
NEED ANSWER ASAP!!!!
gayaneshka [121]

Answer:

y=4x+64

Step-by-step explanation:

The slope intercept form is y=mx+b, m being the slope and b being the y-intercept

The line intersects the y axis at (0, 64), so the y intercept is 64

To find the slope, find the change in y over the change in x

The y decreases by 16 every time the x increases by 4

SO the slope is 16/4, simplified to 4

y=4x+64

6 0
2 years ago
Find the function of which is the solution of 36y"-48y'-48y=0 with initial conditions y1(0)=1. y1'(0)=0
lilavasa [31]

Answer:

y=\frac{1}{4} e^{2x}+\frac{3}{4} e^{-\frac{2}{3}x }

Step-by-step explanation:

Let,  D=\frac{d}{dx}

Now, us simplify the given differential equation and write it in terms of D,

36y''-48y'-48y=0

or, 3y''-4y'-4y=0

or, (3D^2-4D-4)y=0

We have our auxiliary equation:

3D^2-4D-4=0

or, (3D+2)(D-2)=0

or, D=2, - \frac{2}{3}

Therefore our solution is,

y=Ae^{2x}+Be^{- \frac{2}{3}x}

and, y'=2Ae^{2x}-\frac{2}{3}Be^{\frac{2}{3}x }

Applying the boundary conditions, we get,

A+B=1

2A-\frac{2}{3}B=0

Solving them gives us,

A=\frac{1}{4} ,B=\frac{3}{4}

Hence,

y=\frac{1}{4} e^{2x}+\frac{3}{4} e^{-\frac{2}{3}x }

4 0
3 years ago
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