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3 years ago

87 less than the quotient of an unknown number and 43 is -75

Mathematics

2 answers:

slava [35]3 years ago

8 0

The quotient of the unknown number and 43 is 12.
-75 + 87 = 12
12 - 87 = 75

Dafna11 [192]3 years ago

3 0

Hm...maybe searching it up on google can help you...

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Parallel / Perpendicular Practice

deff fn [24]

The slope and intercept form is the form of the straight line equation that includes the value of the slope of the line

Neither
║
Neither
⊥
║
Neither
Neither
Neither

Reason:

The slope and intercept form is the form y = m·x + c

Where;

m = The slope

Two equations are parallel if their slopes are equal

Two equations are perpendicular if the relationship between their slopes, m₁, and m₂ are; $m_1 = -\dfrac{1}{m_2}$

1. The given equations are in the slope and intercept form

$\ y = 3 \cdot x + 1$

The slope, m₁ = 3

$y = \dfrac{1}{3} \cdot x + 1$

The slope, m₂ = $\dfrac{1}{3}$

Therefore, the equations are neither parallel or perpendicular

Neither

2. y = 5·x - 3

10·x - 2·y = 7

The second equation can be rewritten in the slope and intercept form as follows;

$y = 5 \cdot x -\dfrac{7}{2}$

Therefore, the two equations are parallel

3. The given equations are;

-2·x - 4·y = -8

-2·x + 4·y = -8

The given equations in slope and intercept form are;

$y = 2 -\dfrac{1}{2} \cdot x$

Slope, m₁ = $-\dfrac{1}{2}$

$y = \dfrac{1}{2} \cdot x - 2$

Slope, m₂ = $\dfrac{1}{2}$

The slopes

Therefore, m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

The lines are Neither parallel nor perpendicular

Neither

4. The given equations are;

2·y - x = 2

$y = \dfrac{1}{2} \cdot x +1$

m₁ = $\dfrac{1}{2}$

y = -2·x + 4

m₂ = -2

Therefore;

$m_1 \neq -\dfrac{1}{m_2}$

Therefore, the lines are perpendicular

5. The given equations are;

4·y = 3·x + 12

-3·x + 4·y = 2

Which gives;

First equation, $y = \dfrac{3}{4} \cdot x + 3$

Second equation, $y = \dfrac{3}{4} \cdot x + \dfrac{1}{2}$

Therefore, m₁ = m₂, the lines are parallel

6. The given equations are;

8·x - 4·y = 16

Which gives; y = 2·x - 4

5·y - 10 = 3, therefore, y = $\dfrac{13}{5}$

Therefore, the two equations are neither parallel nor perpendicular

Neither

7. The equations are;

2·x + 6·y = -3

Which gives $y = -\dfrac{1}{3} \cdot x - \dfrac{1}{2}$

12·y = 4·x + 20

Which gives

$y = \dfrac{1}{3} \cdot x + \dfrac{5}{3}$

m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

Neither

8. 2·x - 5·y = -3

Which gives; $y = \dfrac{2}{5} \cdot x +\dfrac{3}{5}$

5·x + 27 = 6

$x = -\dfrac{21}{5}$

Therefore, the slopes are not equal, or perpendicular, the correct option is Neither

Learn more here:

brainly.com/question/16732089

6 0

3 years ago

25x2 + 5x

poizon [28]

Answer:

option C: 5x+1

Explaination:

take 5x common and simplify

3 0

2 years ago

Solve the quadratic by factoring. x^2-5x-31=5

torisob [31]

Answer:

$x = - 4\: or \: x = 9$

Step-by-step explanation:

make one side equal to zero:

${x}^{2} - 5x - 36 = 0$

find two factors that multiply to -36 and add to -5 (-9 & 4)

${x }^{2} - 9x + 4x - 36 = 0$

factorise

$x(x - 9) + 4(x - 9) = 0$

$(x + 4)(x - 9) = 0$

solve for x

$x = - 4 \: or \: x = 9$

5 0

3 years ago

The Venn diagram shows the results of two events resulting from rolling a number cube.

gulaghasi [49]

Answer: $P(A\cap B)=\frac{1}{3}$

$P(A)=\frac{1}{3}$

$P(B)=1$

$P(A|B)=\frac{1}{3}$

Step-by-step explanation:

From the given figure it can be seen that

Total number on cube= n(S)=6

Intersection of A and B = n(A ∩ B)= 2

Therefore,

$P(A\cap B)=\frac{n(A\cap B)}{n(s)}=\frac{2}{6}=\frac{1}{3}$

Also, The number of elements in A = 2

Therefore,

$P(A)=\frac{n(A)}{n(s)}=\frac{2}{6}=\frac{1}{3}$

Similarly, The number of elements in B= 6

Therefore,

$P(B)=\frac{n(B)}{n(s)}=\frac{6}{6}=1$

The formula to find the conditional probability is given by :-

$P(A|B)=\frac{P(A\cap B)}{P(B)}\\\\\Rightarrow P(A|B)=\frac{\frac{1}{3}}{1}=\frac{1}{3}$

4 0

3 years ago

Read 2 more answers

Show all work to factor x4 − 10x2 + 9 completely.

leva [86]

Answer:

(x+1)(x-1)(x+3)(x-3)

Step-by-step explanation:

x4-10x^2+9

Group expression so that the coefficients of the x^2 terms add up to +9.

= x^4 -9x^2 - x^2+9

match coefficients in both groups

= x^4 -9x^2 - (x^2-9)

factor each group

= x^2 (x^2-9) - 1(x^2-9)

now factor out the common factor (x^2-9)

= (x^2-1)(x^2-9)

Finally, factor each quadratic factor

= (x+1)(x-1)(x+3)(x-3)

6 0

3 years ago

Read 2 more answers