All categories

3 years ago

14x+7y=217 14x+3y=189

Mathematics

1 answer:

kondor19780726 [428]3 years ago

4 0

Answer:

y = 7

x = 12

Step-by-step explanation:

14x+7y=217, then 14x = 217 - 7y

14x+3y=189, then 14x = 189 - 3y

217 - 7y = 189 - 3y

add 7y to each side of the equation:

4y + 189 = 217

subtract 1890 from each side:

4y = 28

divide both sides by 4:

y = 7

14x + 3(7) = 189

subtract 21 from each side:

14x = 168

divide both sides by 14:

x = 12

You might be interested in

Which statement describes the relationship between x and y

musickatia [10]

The points lie on a line. Such a pattern is called a linear relationship because it represents a straight line relationship between the coordinates of the points. We can describe the relationship between x and y in words as follows: The y-coordinate is three times the x-coordinate.....hope this helps

7 0

3 years ago

Read 2 more answers

SOLVE for x. log3 8 = x

Setler [38]

You are trying so solve
3^x=8
and since 8 is not an integral power of x you need a calculator.
it is ln(8)/ln(3)
or
log(8)/log(3)

5 0

3 years ago

3(x+4) - (x-1) x=3 y=2

Ksenya-84 [330]

Answer: 19

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Attachment below algebra helppppp

yKpoI14uk [10]

Answer: second option.

Step-by-step explanation:

In order to solve this exercise, it is necessary to remember the following properties of logarithms:

$1)\ ln(p)^m=m*ln(p)\\\\2)\ ln(e)=1$

In this case you have the following inequality:

$e^x>14$

So you need to solve for the variable "x".

The steps to do it are below:

1. You need to apply $ln$ to both sides of the inequality:

$ln(e)^x>ln(14)$

2. Now you must apply the properties shown before:

$(x)ln(e)>ln(14)\\\\(x)(1)>2.63906\\\\x>2.63906$

3. Then, rounding to the nearest ten-thousandth, you get:

$x>2.6391$

3 0

3 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago