All categories

3 years ago

Half the difference between the fourth multiple of a number and 16 is 6. What is the number?

Mathematics

1 answer:

Musya8 [376]3 years ago

5 0

Answer:

The number you are looking for is 28.

Step-by-step explanation:

Half the difference of the fourth multiple of a number and 16 equaling 6 looks like this equation:

N - 16/2 = 6.

The first step to finding the value of "N" is to multiply each side by 2.

2(N - 16/2) = 2(6).

The result is N - 16 = 12.

Next, you need to isolate "N" by itself.

Perform inverse operations to get "N" by itself. We need to get rid of -16.

The inverse operation of subtraction is addition.

So add 16 to both sides.

N - 16 + 16 = 12 + 16.

The result is N = 28.

That means the number that makes the problem true is 28.

Let's check to make sure by plugging 28 into the original equation.

28 - 16/2 = 6.

12/2 = 6.

6 = 6.

So the answer to this question is that the number is 28.

You might be interested in

Please could you find the answers to the questions in the attachment.

Fudgin [204]

( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ )we need 3 equations
1. midpoint equation which is ( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ ) when you have 2 points

2. distance formula which is D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height

so

x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=( $\frac{11+19}{2} , \frac{10+6}{2}$ )
midpoint=( $\frac{30}{2} , \frac{16}{2}$ )
midpoint= (15,8)

point x=(15,8)

y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=( $\frac{5+21}{2} , \frac{8+0}{2}$ )
midpoint=( $\frac{26}{2} , \frac{8}{2}$ )
midpoint=(13,4)

Y=(13,4)

legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= $\sqrt{(19-11)^{2}+(6-10)^{2}}$
D= $\sqrt{(8)^{2}+(-4)^{2}}$
D= $\sqrt{64+16}$
D= $\sqrt{80}$
D= $4 \sqrt{5}$
BC= $4 \sqrt{5}$

X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= $\sqrt{(13-15)^{2}+(4-8)^{2}}$
D= $\sqrt{(-2)^{2}+(-4)^{2}}$
D= $\sqrt{4+16}$
D= $\sqrt{20}$
D= $2 \sqrt{5}$
XY= $2 \sqrt{5}$

the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC= $4 \sqrt{5}$ and XY= $2 \sqrt{5}$

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= $\sqrt{(21-5)^{2}+(0-8)^{2}}$
D= $\sqrt{(16)^{2}+(-8)^{2}}$
D= $\sqrt{256+64}$
D= $\sqrt{320}$
D= $4 \sqrt{2}$
AD= $4 \sqrt{2}$

so we have
AD= $4 \sqrt{2}$
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
( $4 \sqrt{2}+4 \sqrt{5}$ ) times 1/2 times $2 \sqrt{5}$ =
( $4 \sqrt{2}+4 \sqrt{5}$ ) times \sqrt{5} [/tex] =
$4 \sqrt{10}+4*5$ = $4 \sqrt{10}+20$ =80 $\sqrt{10}$ =252.982

X=(15,8)
Y=(13,4)
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$
Area=80 $\sqrt{10}$ square unit or 252.982 square units

7 0

3 years ago

A quadratic function is defined by

ella [17]

Answer:

1. The vertex of the graph of the function is (-4, 7)

2. The vertex represents a minimum value

3. The equation represented by the new graph is g(x) = (x + 4)² + 1

Step-by-step explanation:

The given quadratic function is given in vertex form, y = a·(x - h)² + k, as follows;

g(x) = (x + 4)² + 7

1. By comparing the given quadratic function and the vertex form of a quadratic equation, we have;

a = 1, h = -4, and k = 7

The vertex of the graph of the function, (h, k) = (-4, 7)

2. Given that a = 1 > 0, the graph of the quadratic function opens upwards and the vertex represents a minimum value

3. Shifting the graph 6 units down from where it is now will give;

The vertex = (h, k - 6) = (-4, 7 - 6) = (-4, 1)

h = -b/(2·a), k = (-b²/(4·a) + c = -a·h² + c

c = k + a·h²

Therefore, initial value of the constant term, c = -1×(-4)² + 7 = 23

After the shifting the graph 6 units down, c = 23 - 6 = 17

∴ k = 1 = -a·(-4)² + 17

∴ -16/16 = -1 = -a

a = 1

Therefore, the equation represented by the new graph is g(x) = (x + 4)² + 1.

5 0

3 years ago

Can some help me out with this EXTRA POINTS!!

Butoxors [25]

Answer: ≈ 185.73009

Step-by-step explanation:

area of square: $length^{2}$

area of circle: $\frac{radius^{2} * \pi }{y}$

length: 15

radius = diameter/2

radius: 5

area of square - area of circle = area between square and circle:

$15^{2} - \frac{5^{2} * \pi }{2} = 225 - \frac{25\pi}{2}$ ≈ 185.73009

hope it helps!

Please mark as brainliest.

8 0

2 years ago

A figure has side lengths as marked in the diagram. <br> What is the area of the figure?

vovangra [49]

Answer:

132.4

Step-by-step explanation:

7 0

3 years ago

Solve for the variable: -11t + 23 = 56

Annette [7]

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

<span> 3*x+23-(56)=0 </span>

3 0

3 years ago