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ankoles [38]
3 years ago
11

I need help with this problem

Mathematics
1 answer:
sesenic [268]3 years ago
3 0
I hope this helps you

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The equation of a line is y= 4x - 1 and the point (3, p) lies on this line. Determine the value of P
tatyana61 [14]

Answer: 11

Step-by-step explanation:

Given

The equation of the line y=4x-1 passes through the point (3,p)

So, point must satisfies the line i.e.

\Rightarrow p=4(3)-1\\\Rightarrow p=12-1\\\Rightarrow p=11

Therefore, the value of p is 11.

8 0
3 years ago
(12) - (23 +131) =<br> Express your answer in the form (a + bi),
4vir4ik [10]

Answer:

-142

Step-by-step explanation:

First solve parenthesis,

(12) - (23 +131) =

(12) -154

Finally subtract,

12-154

= -142

5 0
3 years ago
Divide and Check...........
kirill [66]

soo this may seem a little awkwark but it still should provide the same resault however its faster :)

so we have

(48x^5-16x^3+40x)/8x

What we are going to be doing is factoring out any and all possibilities for

(48x^5-16x^3+40x)

first factor

(8x(6x^4-2x^2+5))/8x

when we get to this step simplify 8x

we are left with 6x^4-2x^2+5

in order to get an answer that would often be used by long devision just get rid of the +5

This is just a remainder using long devision your instructer may ask for

6x^4-2x^2

Hope it helps

8 0
3 years ago
Read 2 more answers
A car travels at a speed of 85 miles per hour.how far will it travel in 15minutes?
Eduardwww [97]

Answer:

21.25

Step-by-step explanation:

5 0
3 years ago
How do you solve 1/x + 1/y + 1/z = 1 for z?
garri49 [273]

Answer:

\large\boxed{z=\dfrac{xy}{xy-x-y}}

Step-by-step explanation:

\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\qquad\text{multiply both sides by}\ xyz\neq0\\\\xyz\cdot\dfrac{1}{x}+xyz\cdot\dfrac{1}{y}+xyz\cdot\dfrac{1}{z}=xyz\cdot1\qquad\text{simplify}\\\\yz+xz+xy=xyz\qquad\text{subtract}\ xy\ \text{from both sides}\\\\xz+yz=xyz-xy\qquad\text{subtract}\ xyz\ \text{from both sides}\\\\xz+yz-xyz=-xy\qquad\text{distribute}\\\\(x+y-xy)z=-xy\qquad\text{divide both sides by}\ (x+y-xy)\\\\z=\dfrac{-xy}{x+y-xy}\\\\z=\dfrac{-xy}{-(xy-x-y)}\\\\z=\dfrac{xy}{xy-x-y}

5 0
3 years ago
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