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2 years ago

Solve for x (picture is included below)

Mathematics

2 answers:

Anna11 [10]2 years ago

8 0

Solve for x:

4x - 7 = 3

Add 7 to both sides

4x = 10

Divide both sides by 4

x = 10/4

Simplify:

x = 5/2

Answer for question 1: x = 5/2

13 + 2x/3 = 15

Multiply both sides by 3

39 + 2x = 45

Subtract 39 from both sides

2x = 6

Divide both sides by 2

x = 3

Answer for question 2: x = 3

10x + 7 = 15

Subtract 7 from both sides

10x = 8

Divide both sides by 10

x = 8/10

Simplify

x = 4/5

Answer for question 3: x = 4/5

zlopas [31]2 years ago

4 0

Step-by-step explanation:

We have to find the value of x.

1.) 4x - 7 = 3

Add the like terms

4x = 3 + 7

Subtract 7 from each section

4x - 7 = 3

4x = 3 + 7

4x = 10

x = 10/4

x = 5/2

2.) 13 + 2x/3 = 15

(39 + 2x)/3 = 15

39 + 2x = 15 × 3

39 + 2x = 45

Subtract 45 from each term.

39 - 45 + 2x = 45 - 45

- 6 + 2x = 0

2x = 6

x = 6/2

x = 3

3.) 10x + 7 = 15

Subtract 15 from each terms

10x + 7 - 15 = 15 - 15

10x - 8 = 0

10x = 8

x = 8/10

x = 4/5

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Please help Find the area in square units of ABC△ plotted below.

neonofarm [45]

Answer:

Use the distance formula on both points AC and AB.

Distance formula is this:

\begin{gathered}d=\sqrt{(x2-x1)^2+(y2-y1)^2} \\\\d=\sqrt{(1--5)^2+(8--7)^2} \\\\d=\sqrt{(6)^2+(15)^2} \\\\d=\sqrt{36+225} \\\\d=\sqrt{261} \\\\\end{gathered}d=(x2−x1)2+(y2−y1)2d=(1−−5)2+(8−−7)2d=(6)2+(15)2d=36+225d=261

Distance for AC is 16.16

Now do the same with the numbers for AB and get the distance of 5.39

2. To get the area, use the formula 1/2 x base x height

AB is the base and AC is the height.

1/2 x 16.16 x 5.39 = 43.55

the answer is 43.5

6 0

1 year ago

Solve for

Varvara68 [4.7K]

Answer:

t = 1, but please check to see if I correctly interpreted your formatting.

Step-by-step explanation:

The format of the question is too confusing.

I'll assume this is the equation to be solved:

16−2t=t+9+4t

16−2t=t+9+4t

16-9 = t + 2t + 4t

7 = 7t

t = 1

7 0

1 year ago

Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p

Juli2301 [7.4K]

I'm reading this as

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy$

with $\nabla f=(2xe^{-y},2y-x^2e^{-y})$ .

The value of the integral will be independent of the path if we can find a function $f(x,y)$ that satisfies the gradient equation above.

You have

$\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}$

Integrate $\dfrac{\partial f}{\partial x}$ with respect to $x$ . You get

$\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx$
$f=x^2e^{-y}+g(y)$

Differentiate with respect to $y$ . You get

$\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]$
$2y-x^2e^{-y}=-x^2e^{-y}+g'(y)$
$2y=g'(y)$

Integrate both sides with respect to $y$ to arrive at

$\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy$
$y^2=g(y)+C$
$g(y)=y^2+C$

So you have

$f(x,y)=x^2e^{-y}+y^2+C$

The gradient is continuous for all $x,y$ , so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e$