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1 year ago

Four more than the product of a number and 3 equals 7.

1 answer:

7 0

The phrase you typed can be written in mathematical form using a replacement of "a number" with the letter "x" (for our unknown).

So we can translate in mathematical form as shown below:

"<em>four more than the product of a number and 3 equals 7</em>"

x * 3 + 4 = 7

and of course we can solve this equation by operating with thenumbers so as to isolate the unknown "x"on one side of the equal sign:

Subtract 4 from both sides:

x * 3 = 7 - 4

x * 3 = 3

Divide both sides by 3 (so as to eliminate the factor 3 multiplying x):

x = 3 / 3

x = 1

And our answer is that the unknown number is 1 (one).

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Solve by substitution x-y=-11 y+7=-2x

Bess [88]

x - y = - 11 .............( 1 )

y + 7 = - 2x ...............( 2 )

from equation ( 1 )

x - y = - 11

x = -11 + y ...........( 3)

putting x in equation ( 2 )

y + 7 = - 2 x

y + 7 = -2 ( -11 + y ) y + 7 = 22 - 2 y y + 2 y =22 - 7

3 y = 15 y = 15 / 3

putting value of y in equation 3

x = -11+ ( 13 /5 ) x = -33 /3 + 15 / 3 ( l.c.m)

x = -17 / 5

check

x = -11 + y

- 17 / 5 = -11 + 13 / 5

-17 /5 = -17 / 5

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3 years ago

8X over five -6 equals 2X +4 how do you solve that

yaroslaw [1]

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3 years ago

In an RPG game, each player can choose among 5 basic classes: Mage, Warrior, Archers, Priests, Monks. Each basic class can advan

grigory [225]

Answer:

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Step-by-step explanation:

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2 years ago

Solve the following quadratic function<br> by utilizing the square root method.<br> y = 16 – x2

xxMikexx [17]

Answer:

The solutions are x = -4 and x = 4.

Step-by-step explanation:

Solving a quadratic equation:

We have to find x for which $y = 0$ .

In this question:

$y = 16 - x^2$

$16 - x^2 = 0$

$x^2 = 16$

$x = \pm \sqrt{16}$

$x = \pm 4$

The solutions are x = -4 and x = 4.

3 0

3 years ago

2. Consider the circle x^2−4x+y^2+10y+13=0. There are two lines tangent to this circle having a slope of 2/3.

likoan [24]

Answer:

a) The coordinates are (0.431, -2.646) and (3.568,-7.354)

b) The tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

Step-by-step explanation:

First, lets complete squares, by taking for each cordinate the square of a linear expression

0= x²−4x+y²+10y+13 = (x-2)²+ 4 + (y+5)²-25+13 = (x-2)²+(y+5)² - 8

Hence (x-2)² + (y+5)² = 8

Lets put y in function of x. We should obtain 2 functions f and g that represent the circle.

(y+5)² = 8 - (x-2)² = -x² + 4x + 4

$y = ^+_- \sqrt{-x^2+4x+4} -5$

Thus

$f(x) = \sqrt{-x^2+4x+4} - 5$

$g(x) = -\sqrt{-x^2+4x+4} - 5$

Lets find the derivate of each function and the points in which they reach the value 2/3. In those points the tangent line will have a slope of 2/3.

We may just find the values of x which the derivate of f is either 2/3 or -2/3.

$\frac{-x+2}{\sqrt{-x^2+4x+4}} = ^+_-\frac{2}{3} \Leftrightarrow \frac{x^2-4x+4}{-x^2+4x+4} = \frac{4}{9} \Leftrightarrow 9(x^2-4x+4) = 4(-x^2+4x+4) \\\Leftrightarrow 13x^2-52x+20 = 0$

The quadratic has roots

$\frac{52 ^+_-\sqrt{1664}}{26}$

one root is 3.568, which corresponds with g, and the other root is 0.431, corresponding to f.

Also g(3.568) = - 7.354 and f(0.431) = -2.646

This means that the coordinates are (0.431 , -2.646), (3.568 , -7.354) and the tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

5 0

2 years ago