All categories

2 years ago

A number - _______ = previous number

Mathematics

1 answer:

leva [86]2 years ago

6 0

Answer:

Step-by-step explanation:

When we substract 1 from any number then will get just previous number.

Example :

Let any number 17

17-1 =16

16 is the just previous number of 17

Therefore, A number -1 = previous number

Answer will be 1

You might be interested in

PLEASE HELP!!

Musya8 [376]

Answer:

x = - 33

Step-by-step explanation:

Given

$\frac{2}{3}$ (x + 6) = - 18

Multiply both sides by 3 to clear the fraction

2(x + 6) = - 54 ( divide both sides by 2 )

x + 6 = - 27 ( subtract 6 from both sides )

x = - 33

6 0

3 years ago

Read 2 more answers

Calculate the average speed (tin km/hr) of charlie who run to the store 4 km away in 30 minutes

Galina-37 [17]

ANSWER

$8km/hr$

EXPLANATION

The average speed

$= \frac{distance}{time}$

Distance=4km

Time=30mins

Convert time to hours.

Time =0.5hrs

The average speed

$= \frac{4}{0.5}$

$= 8km/hr$

8 0

3 years ago

Match each equation with its solution set. Tiles a2 − 9a + 14 = 0 a2 + 9a + 14 = 0 a2 + 3a − 10 = 0 a2 + 5a − 14 = 0 a2 − 5a − 1

sattari [20]

We have that

N 1)
a² − 9a + 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 9a)=-14

Complete the square Remember to balance the equation by adding the same constants to each side

(a² − 9a+20.25)=-14+20.25

Rewrite as perfect squares

(a-4.5)²=6.25--------> (a-4.5)=(+/-)√6.25

a1=4.5+√6.25-----> a1=7

a2=4.5-√6.25-----> a2=2

the solution problem N 1 is the pair {7, 2}

N 2)

a² + 9a + 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 9a)=-14

Complete the square Remember to balance the equation by adding the same constants to each side

(a² +9a+20.25)=-14+20.25

Rewrite as perfect squares

(a+4.5)²=6.25--------> (a+4.5)=(+/-)√6.25

a1=-4.5+√6.25-----> a1=-2

a2=-4.5-√6.25-----> a2=-7

the solution problem N 2 is the pair {-2,-7}

N 3)

a² + 3a − 10 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 3a)=10

Complete the square Remember to balance the equation by adding the same constants to each side

(a² + 3a+2.25)=10+2.25

Rewrite as perfect squares

(a+1.5)²=12.25------> (a+1.5)=(+/-)√12.25

a1=-1.5+√12.25-----> a1=2

a2=-1.5-√12.25-----> a2=-5

the solution problem N 3 is the pair {2, -5}

N 4)

a² + 5a − 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 5a) =14

Complete the square Remember to balance the equation by adding the same constants to each side

(a² + 5a+6.25) =14+6.25

Rewrite as perfect squares

(a+2.5)² =20.25-------> (a+2.5)=(+/-)√20.25

a1=-2.5+√20.25-----> a1=2

a2=-2.5-√20.25-----> a2=-7

the solution problem N 4 is the pair {2, -7}

N 5)

a² − 5a − 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 5a)=14

Complete the square Remember to balance the equation by adding the same constants to each side

(a² − 5a+6.25)=14+6.25

Rewrite as perfect squares

(a-2.5)²=2025--------> (a-2.5)=(+/-)√20.25

a1=2.5+√20.25-----> a1=7

a2=2.5-√20.25-----> a2=-2

the solution problem N 5 is the pair {7, -2}

3 0

2 years ago

Read 2 more answers

What is the equation of the quadratic graph with a focus of (1, 1) and a directrix of y = −1? PLs HELP!!

Delicious77 [7]

ANSWER
$f(x)= \frac{1}{4} {(x - 1)}^{2}$

EXPLANATION

Since the directrix is
$y = - 1$

the axis of symmetry of the parabola is parallel to the y-axis.

Again, the focus being,

$(1,1)$

also means that the parabola will open upwards.

The equation of parabola with such properties is given by,

${(x - h)}^{2} = 4p(y - k)$
where
$(h,k)$

is the vertex of the parabola.

The directrix and the axis of symmetry of the parabola will intersect at
$(1, - 1)$

The vertex is the midpoint of the focus and the point of intersection of the axis of the parabola and the directrix.

This implies that,
$h = \frac{1 + 1}{2} = 1$

and
$k = \frac{ - 1 + 1}{2} = 0$

The equation of the parabola now becomes,

$(x - 1) ^{2} = 4p(y - 0)$

$|p| = 1$

Thus, the distance between the vertex and the directrix.

This means that,

$p = - 1 \: or \: 1$

Since the parabola opens up, we choose
$p = 1$
Our equation now becomes,

${(x - 1)}^{2} = 4(1)(y - 0)$

This simplifies to
${(x - 1)}^{2} = 4y$

or

$y = \frac{1}{4} {(x - 1)}^{2}$

This is the same as,

$f(x)= \frac{1}{4} {(x - 1)}^{2}$

The correct answer is D .

4 0

3 years ago

Please help ASAP!!!!

LuckyWell [14K]

Answer:

Step-by-step explanation:

(-7x^2 - 3) + (4x - 6)

-3 - 6 = -9

-7x^2 + 4x - 9

5 0

2 years ago

Read 2 more answers