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2 years ago

Can you help me find x? x/7+x/3=9/7

Mathematics

1 answer:

castortr0y [4]2 years ago

6 0

x/7 - x/3 = 9/7

x/7(3) - x/3(7) = 9/7(3)

3x/21 - 7x/21 = 27/21

3x - 7x = 27

-4x = 27

(1/-4) -4x = 27(1/-4)

x = -27/4

You might be interested in

Complete the table and find the balance A if $5000 is invested an an annual rate of 6% for 15 years and compounded n times a yea

Olenka [21]

Answer:

1: 11982.79

2: 12136.31

4: 12216.09

12: 12270.46

365: 12297.10

Step-by-step explanation:

The compound interest formula is A=P*(1+r/n)^nt

P=principal amount

r=rate

n=number of compounds per period

t=number of periods

In this situation, the principal amount is 5,000=P. The rate is 6%, so r=0.06. n is equal to the top number in the table. And this is 15 years, so t=15.

For 1, it would be A=5,000*(1+.06/1)^1*15, or A=5,000(1.06^15), or 11982.79.

For 2, it would be 5,000*(1+.06/2)^2*15, or 5,000*(1.03^30), or 12136.31.

And so forth. Good luck

5 0

3 years ago

using approximations to 1 significant figure, estimate the value of 9.94² × 29.7 you must show your working

zvonat [6]

Answer:

2934.5

Step-by-step explanation:

2934.46692 Rounded to 1 Sig Fig is 2934.5

4 0

3 years ago

The mean of 5 numbers is ten I add one now the mean is 11 which number did I add?

Leno4ka [110]

Answer:

you added 6

Step-by-step explanation:

how the number order goes.

3 0

2 years ago

What is the area of the trapezoid?

Maslowich

<h3>Given Information :</h3>

Length of parallel sides = 60 ft and 40 ft
Height of the trapezoid = 30 ft

<h3>To calculate :</h3>

Area of the trapezoid.

<h3>Calculation :</h3>

As we know that,

$\bigstar \: \boxed{\sf {Area_{(Trapezium)} = \dfrac{1}{2} \times ( a + b) \times h}} \\$

a and b are length of parallel sides.
h denotes height.

Substituting values, we get :

$\longmapsto$ Area = $\sf \dfrac{1}{2}$ × ( 60 + 40 ) × 30 ft $\sf ^2$

$\longmapsto$ Area = $\sf \dfrac{1}{2}$ × 100 × 30 ft $\sf ^2$

$\longmapsto$ Area = 1 × 100 × 15 ft $\sf ^2$

$\longmapsto$ Area = 100 × 15 ft $\sf ^2$

$\longmapsto$ Area = 1500 ft $\sf ^2$

Therefore,

Area of the trapezoid is 1500 ft $\sf ^2$

6 0

2 years ago

An arhitect wants to draw a rectangle with a diagonal of 20 inches. The length of the rectangle is to be 8 inches more than twic

kiruha [24]

The diagonal of a rectangle = sqrt(w^2 + l^2)

w = width

l = length

In this problem,

The diagonal = 20 in

w = x

l = 2x + 8

Let's plug our numbers into the formula above.

20in = sqrt((x)^2 + (2x + 8)^2)

Let's simplify the inside of the sqrt

20 in = sqrt(5x^2 + 32x + 64)

Now, let's square both sides.

400 = 5x^2 + 32x + 64

Subtract 400 from both sides.

0 = 5x^2 + 32x - 336

Factor

0 = (5x - 28)(x + 12)

Set both terms equal to zero and solve.

x + 12 = 0

Subtract 12 from both sides.

x = -12

5x - 28 = 0

Add 28 to both sides.

5x = 28

Divide both sides by 5

x = 28/5

The width cant be a negative number so now we know that the only real solution is 28/5

Let's plug 28/5 into our length equation.

Length = 2(28/5) + 8 = 56/5 + 8 = 96/5

In conclusion,

Length = 96/5 inches

Width = 28/5

4 0

2 years ago