Find the value of P that makes the equation true

1 answer:

4 0

There are six p's, which add up to give a sum of thirty six

6p=36

Divide six by both sides to get the value of p

36/6 is 6

p is equal to six

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Find a unit vector that has the same direction as the vector a and the opposite direction of the vector

nasty-shy [4]

The unit vector is given by the following formula:
a '= (a) / (lal)
Where,
a: vector a
lal: Vector module a
We are looking for the module:
lal = root ((- 15) ^ 2 + (8) ^ 2)
lal = 17
Same direction:
a = -15i + 8j
The unit vector is:
a '= (1/17) * (- 15i + 8j)
Opposite direction:
a = 15i - 8j
The unit vector is:
a '= (1/17) * (15i - 8j)
Answer:
a unit vector that has the same direction as the vector a is:
a '= (1/17) * (- 15i + 8j)
a unit vector that has the opposite direction of the vector a is:
a '= (1/17) * (15i - 8j)

5 0

3 years ago

What is the value of y for the parallelogram shown?

dybincka [34]

Opposite angles in a parallelogram have to be equal so

3y=66
y=22

3 0

3 years ago

Read 2 more answers

I dont know how to do this without the y intercept

Elza [17]

I think the y intercept is 3.5

7 0

3 years ago

Read 2 more answers

Using a table of values, what is the approximate positive solution to the equation to the nearest quarter of a unit? f(x)=log(x+

torisob [31]

Answer:

The answer is 0.25 for plato/edmentum users

Step-by-step explanation:

The x-intercept/zero for function f is x= -1 and the x-intercept/zero for function g is x= 0, therefore, the closest positive value out of the options is 0.25 sense the problem wants the functions to equal around the same value. Tbh I suck at math and idek if my explanation is right I really just guessed on the problem. Here's a ss for proof that my answer's right tho lol p.s. if ur 15-19 amos: kenz8124 Have a great day :)

6 0

3 years ago

X^3+7x^2+13x+4=0 Show all work

Softa [21]

$\displaystyle\\
x^3+7x^2+13x+4=0\\\\
x^3+\underbrace{4x^2+3x^2}_{7x^2}+\underbrace{12x+x}_{13x}+4=0\\\\
(x^3+4x^2) + (3x^2 + 12x)+(x+4)=0\\\\
x^2(x+4)+3x(x+4)+(x+4)=0\\\\
(x+4)(x^2 + 3x +1)=0\\\\
x+4 = 0~~~\Longrightarrow~~~\boxed{x_1 = -4}\\\\
x^2 + 3x +1=0\\\\
x_{23}= \frac{-b\pm \sqrt{b^2-4ac} }{2a}= \frac{-3\pm \sqrt{3^2-4\cdot 1 \cdot 1} }{2\cdot 1}= \frac{-3\pm \sqrt{5} }{2}\\\\
\boxed{x_2 = \frac{-3-\sqrt{5} }{2}}\\\\
\boxed{x_3 = \frac{-3+\sqrt{5} }{2}}$

4 0

3 years ago