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3 years ago

Pls help me I’m so lost

Mathematics

1 answer:

mojhsa [17]3 years ago

6 0

Answer:

Step-by-step explanation:

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Anyone know how to solve this?? Number 5!!

iragen [17]

P=6+1+2
6+1+2=9 x 3= 27

8 0

3 years ago

25, 8, 10, 35, 5, 45, 40, 30, 20 What is the upper quartile of the data set above?A.35

Vlada [557]

<span>An upper quartile is the range of numbers above the median in a set. Thus, the numbers have to be rearranged mentally or on paper. Fortunately, there are an odd set of numbers, so one number, 25, is the median. The upper quartile is 30, 35, 40, 45.</span>

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3 years ago

i think my my mom figured out i was bi and she is gonna be mad what do i do i think she is gonna whip me HELP ME PLS

Eva8 [605]

If you can you should go to a friends house for a while until you feel like it’s safe to go back

4 0

3 years ago

Read 2 more answers

Tracy sells water at a fair in 17oz bottles or 11oz cans. If on one shelf she has 11 containers of water that hold a total of 15

blondinia [14]

Answer:

Number of bottles = 6

Number of cans = c

Step-by-step explanation:

Given that:

Container for selling water :

17 oz bottle

11 oz can

Let number of bottles = b

Number of cans = c

b + c = 11 - - - (1)

17b + 11c = 157 - - - (2)

b = 11 - c

Substitute into (2)

17(11 - c) + 11c = 157

187 - 17c + 11c = 157

187 - 6c = 157

-6c = 157 - 187

-6c = -30

c = 30/6

c = 5

From ; b = 11 - c

b = 11 - 5

b = 6

Hence,

Number of bottles = 6

Number of cans = c

3 0

3 years ago

1) Determine the discriminant of the 2nd degree equation below:

Aleksandr-060686 [28]

$\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}$

We have, Discriminant formula for finding roots:

$\large{ \boxed{ \rm{x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }}}$

Here,

x is the root of the equation.
a is the coefficient of x^2
b is the coefficient of x
c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5\pm \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm \sqrt{25 - 24} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm 1}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 2 \: or - 3}}}$

❒ p(x) = x^2 + 2x + 1 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{ {2}^{2} - 4 \times 1 \times 1} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm 0}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 1 \: or \: - 1}}}$

❒ p(x) = x^2 - x - 20 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{1 \pm 9}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \: - 4}}}$

❒ p(x) = x^2 - 3x - 4 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm \sqrt{9 + 16} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm 5}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \: - 1}}}$

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5 0

3 years ago

Read 2 more answers