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Juliette [100K]
4 years ago
7

What is (-4,2), slope = 1/2

Mathematics
1 answer:
Rashid [163]4 years ago
4 0
Y1-y=m(x1-x)
Y-2=1/2(x-1)
Y-2=1/2(x+4)
Y-2=1/2x+2
Y=1/2x+4
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A vehicle is travelling from rest. After 10 seconds its velocity will be 20ms find acceleration?​
Paladinen [302]

Initial velocity (u) = 0m/s

Final velocity (v) = 20m/s

Time (t) = 10 s

Acceleration (a)

= (v - u)/t

= [(20m/s) - (0m/s)]/10s

= (20m/s)/10s

= (20m/s²)/10

=> 2m/s²

8 0
3 years ago
The line x - y = 5 passes through which set (-5, 0) (0, 5) (0, -5)
s2008m [1.1K]
The line x - y = 5 passes through points (-5,0) and (0,5)
the line in slope intercept form is y = x + 5 and plug in the point to see if the equation is true.
6 0
4 years ago
Read 2 more answers
What is the slope of 3,5 and -3,-5
olya-2409 [2.1K]
The slope is 5/3 five over 3. You youse the formula y2 - y1 over x2-x1. Plug into calculator,and then reduce.
5 0
4 years ago
Simplify each only using positive exponents:<br> 2x^-3 • 4x^2<br> 2x^4 • 4x^-3<br> 2x^3y^-3 • 2x
erica [24]
<h2>Answer:</h2>

\frac{2}{x}

\frac{x}{2}

\frac{4x^4}{y^3}

<h2>Step-by-step explanation:</h2>

a. 2x^-3 • 4x^2

To solve this using only positive exponents, follow these steps:

i. Rewrite the expression in a clearer form

2x⁻³ . 4x²

ii. The position of the term with negative exponent is changed from denominator to numerator or numerator to denominator depending on its initial position. If it is at the numerator, it is moved to the denominator. If otherwise it is at the denominator, it is moved to the numerator. When this is done, the negative exponent is changed to positive.

In our case, the first term has a negative exponent and it is at the numerator. We therefore move it to the denominator and change the negative exponent to  positive as follows;

\frac{1}{2x^3} . 4x^2

iii. We then solve the result as follows;

\frac{1}{2x^3} . 4x^2 = \frac{2}{x}

Therefore, 2x⁻³ . 4x² = \frac{2}{x}

b. 2x^4 • 4x^-3

i. Rewrite as follows;

2x⁴ . 4x⁻³

ii. The second term has a negative exponent, therefore swap its position and change the negative exponent to a positive one.

2x^4 . \frac{1}{4x^3}

iii. Now solve by cancelling out common terms in the numerator and denominator. So we have;

\frac{x}{2}

Therefore, 2x⁴ . 4x⁻³ = \frac{x}{2}

c. 2x^3y^-3 • 2x

i. Rewrite as follows;

2x³y⁻³ . 2x

ii. Change position of terms with negative exponents;

2x^3.\frac{1}{y^3} .2x

iii. Now solve;

\frac{4x^4}{y^3}

Therefore, 2x³y⁻³ . 2x = \frac{4x^4}{y^3}

8 0
3 years ago
Help me I'm stuck!!!
Alchen [17]
Volume of the tank = length x width x height
Volume = 15.4 x 10 x 3 =  450 in³

Volume of one gallon = 231 in³

Then we need 450/231 gallons to fill the tank, that is 450/231 = 1.95 gallons
8 0
3 years ago
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