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3 years ago

Can anyone plz help me

Mathematics

1 answer:

Andrew [12]3 years ago

6 0

Let's solve number 7 step-by-step.
y/23=7
Step 1: Multiply both sides by 23.
y/23=7
(y23)*(23)=(7)*(23)
y=161 is our answer.

Let's solve number 8 step-by-step.
n−4.85=12.6
Step 1: Add 4.85 to both sides.
n−4.85+4.85=12.6+4.85
n=17.45 is our answer.


Let's solve number 12 step-by-step.
m/5=8
Step 1: Multiply both sides by 5.
m/5=8
(m/5)*(5)=(8)*(5)
m=40 is our answer.


Let's solve number 16 step-by-step.
a/12=8
Step 1: Multiply both sides by 12.
a/12=8
(a/12)*(12)=(8)*(12)
a=96 is or answer.

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the answer would be 408

just set it up like this

12 48

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3 years ago

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

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

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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Answer:

The answer is 20

Step-by-step explanation:

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Geometry help please! Match the reasons with the statements

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Given ⇔ 1. ∠PRS and ∠VUW are supplementary
Angles forming a linear pair sum of 180° ⇔ 3. ∠PRS + ∠SRU = 180°
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Step-by-step explanation:

Here we have , ∠PRS and ∠VUW are supplementary . We need to complete the proof of TV || QS , with matching the reasons with statements .Let's do this :

Given ⇔ 1. ∠PRS and ∠VUW are supplementary
Angles forming a linear pair sum of 180° ⇔ 3. ∠PRS + ∠SRU = 180°
Definition of Supplementary angle ⇔ 2. ∠PRS + ∠VUW = 180°
Transitive property of equality ⇔ 4 . ∠PRS + ∠VUW = ∠PRS + ∠SRU
Algebra ⇔ 5. ∠VUW = ∠SRU
Converse of Corresponding angle Postulate ⇔ Line TV || Line QS

Above mentioned are , are the statements matched with expressions on right hand side (RHS) .

The Corresponding Angles Postulate states that, when two parallel lines are cut by a transversal , the resulting corresponding angles are congruent .

The converse states: If corresponding angles are congruent, then the lines cut by the transversal are parallel.

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