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borishaifa [10]
2 years ago
8

Guys i need help on this problem I'm stuck

Mathematics
2 answers:
Alex73 [517]2 years ago
7 0
The answer is y = mx + b
m is the slope and b is the y-intercept
Norma-Jean [14]2 years ago
6 0

Answer:

y=mx+b is slope intercept form

Step-by-step explanation:

letter B :)

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PLZZZZZZZZZZZZZZZZZZZZZZZZZ HELP!!!
NISA [10]

Answer:

Step-by-step explanation:

Instead of dividing both sides by 5, it is multiplied

5x + 3 = 6            {Subtract 3 from both sides}

5x + 3 - 3 = 6-3

            5x = 3       {Divide both sides by 5}

           5x/5 = 3/5

                 x = 3/5

8 0
2 years ago
Solve for x: (2/5)(x-4)=2x​
NARA [144]

Simplify brackets

2/5(x - 4) = 2x

Simplify 2/5(x - 4) to 2(x - 4)/5

2(x - 4)/5 = 2x

Multiply both sides by 5

2(x - 4) = 10x

Divide both sides by 2

x - 4 = 5x

Subtract x from both sides

-4 = 5x - x

Simplify 5x - x to 4x

-4 = 4x

Divide both sides by 4

-1 = x

Switch sides

<u>x = -1</u>

8 0
3 years ago
Find the area of the<br> parallegram
sweet [91]

Answer:

To find the area, multiply the base by the height. The formula is: A = B * H where B is the base, H is the height, and * means multiply. The base and height of a parallelogram must be perpendicular.

Step-by-step explanation:

hope it helps

please mark brainliest

3 0
2 years ago
How do i Simplify: b^8 b^4
11Alexandr11 [23.1K]
It  looks like they're multiplied, then you can simply add the exponents,

b⁸ * b⁴ = b⁸⁺⁴ = b¹²

remember, b⁸=b*b*b*b*b*b*b*b and b⁴=b*b*b*b

so b⁸ * b⁴ = b*b*b*b*b*b*b*b * b*b*b*b = b¹²
7 0
3 years ago
The line L is a tangent to the curve with equation y= 4x^2 +1 . The line L cuts the y axis at (0,8) and has a positive gradient.
sergeinik [125]

A generic point on the graph of the curve has coordinates

(x, 4x^2+1)

The derivative gives us the slope of the tangent line at a given point:

f(x) = 4x^2+1 \implies f'(x) = 8x

Let k be a generic x-coordinate. The tangent line to the curve at this point will pass through (k, 4k^2+1) and have slope 8k

So, we can write its equation using the point-slope formula: a line with slope m passing through (x_0, y_0) has equation

y-y_0 = m(x-x_0)

In this case, (x_0, y_0)=(k, 4k^2+1) and m=8k, so the equation becomes

y-4k^2-1 = 8k(x-k)

We can rewrite the equation as follows:

y-4k^2-1 = 8k(x-k) \iff y = 8kx - 8k^2+4k^2+1 \iff y = 8kx-4k^2+1

We know that this function must give 0 when evaluated at x=0:

f(x) = 8kx-4k^2+1 \implies f(0) = -4k^2+1 = 8 \iff -4k^2 = 7 \iff k^2 = -\dfrac{7}{4}

This equation has no real solution, so the problem looks impossible.

5 0
3 years ago
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