All categories

3 years ago

If you solve this for me I’ll love you forever (but seriously I need the answer ASAP)

Mathematics

2 answers:

mash [69]3 years ago

8 0

Description:

This will be no solution because when we add 10 to both sides of the equation then simplify it. After that we will subtract 6 from both sides. It will give us no solution. For more info please see the attachment.

Answer: No solution

Hope this helps.

Anestetic [448]3 years ago

5 0

Answer:

Step-by-step explanation:

A variable represents an actual value. We may not know what value it represents, but it must represent some number.

Look at what we are given. 2(3x-5) = 6x + 8. The next step to solving this equation is using the distributive property of multiplication over addition. If you need help understanding this property, I recommend you either create a new question or ask me about it in the comments.

In any case, multiplying 2 and (3x-5) equals by the distributive property 6x -10.

So 6x-10 = 6x+8

If this is true, we can subtract 6x from both sides. If two things are equal, subtracting the same value from both sides will give two new things that are also equal.

In this case, subtracting 6x gives -10 = 8.

But -10 does not equal 8. So what does that mean? It basically means, that whatever value x represents, 6x-10 will not equal 6x+8. And if that's true, then no matter what x is, 2(3x-5) will not equal 6x+8. Not a single solution for x could satisfy this equation, so we can say that there is no solution.

You might be interested in

Consider f and c below. f(x, y) = (3 + 4xy2)i + 4x2yj, c is the arc of the hyperbola y = 1/x from (1, 1) to 3, 1 3 (a) find a fu

irga5000 [103]

We want to find a scalar function $f(x,y)$ such that $\nabla f(x,y)=\mathbf f(x,y)=\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j$ .
So we need to have
$\dfrac{\partial f}{\partial x}=3+4xy^2$
Integrating both sides with respect to $x$ gives
$f(x,y)=3x+2x^2y^2+g(y)$
Differentiating with respect to $y$ gives
$\dfrac{\partial f}{\partial y}=4x^2y+\dfrac{\mathrm dg}{\mathrm dy}=4x^2y$ $\implies\dfrac{\mathrm dg}{\mathrm dy}=0$ $\implies g(y)=C$
So we find that
$f(x,y)=3x+2x^2y^2+C$
By the fundamental theorem of calculus, we then know the line integral depends only on the values of $f(x,y)$ at the endpoints of the path. Therefore

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f\left(3,\frac13\right)-f(1,1)=11-5=6$

4 0

3 years ago

Can someone please help me with this problem, it is really confusing

levacccp [35]

Answer =-0.7x +0

Find the gradient by y2-y1/x2-x1
-14 -7/ 20- -10
-21/30
-0.7
And plug in the coordinates to find the y intercept
7= -0.7 (-10) + c
C=0

4 0

2 years ago

Use >,<,=. 0.94___0.49

uranmaximum [27]

0.94>0.49 because 0.94 is the larger number.

5 0

3 years ago

Read 2 more answers

Which of the following graphs represents a function

Nata [24]

Where are the graphs? a function is when every input is paired with exactly one output(every x value must have exactly 1 y value). the vertical line test helps determine a function, a vertical line can’t intersect a graph at more then one point, if it does intersect in more then 1 point then it’s not a function

7 0

3 years ago

Multiply Conjugates (r+1/4)r-1/4)

Mamont248 [21]

Answer:

$\large\boxed{\left(r+\dfrac{1}{4}\right)\left(r-\dfrac{1}{4}\right)=r^2-\dfrac{1}{16}}$

Step-by-step explanation:

$\text{Use}\ (a+b)(a-b)=a^2-b^2\\\\\left(r+\dfrac{1}{4}\right)\left(r-\dfrac{1}{4}\right)=r^2-\left(\dfrac{1}{4}\right)^2=r^2-\dfrac{1^2}{4^2}=r^2-\dfrac{1}{16}$

8 0

3 years ago

Read 2 more answers