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3 years ago

Help solve for x you ant gotta solve for the other one

Mathematics

2 answers:

vladimir1956 [14]3 years ago

6 0

Answer:

could you maybe specify a little bit more

velikii [3]3 years ago

5 0

Answer:

hi its not clear

Step-by-step explanation:

you can send onother photho

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Simplify this expression

Juliette [100K]

x^2+2x+1
= x^2 + 1x + 1x + 1
= (x^2+1x) + (1x+1)
= (x+1)(x+1)

3 0

3 years ago

Help, with solving rational equations! I am no good at it :( Solve for x: 2 over 5 plus 3 over 5 x equals quantity x plus 5 over

mamaluj [8]

If this is your equation: $\frac{2}{5} + \frac{3}{5x} = \frac{x+5}{10}$
Solution:
LCD for 5 and 5x is 5x
$\frac{x}{x} ( \frac{2}{5} )+ \frac{3}{5x} = \frac{x+5}{10}$
$\frac{2x}{5x} + \frac{3}{5x} = \frac{x+5}{10}$
$\frac{2x+3}{5x} = \frac{x+5}{10}$
$10(2x+3)=5x(x+5)$ ←cross product
20x + 30 = 5x² + 25x ←simplify with distributive property
0 = 5x² + 5x - 30 ←use inverse operations to collect all terms on one side
0 = x² + x - 6 ←if possible divide by numerical GCF (This case 5)
0 = (x + 3)(x - 2) ←Factor
x = -3 or x = 2
Please check by substitution in original equation... Both work

4 0

3 years ago

Find the constant of proportionality k. Then write an equation for the relationship between x and y

ExtremeBDS [4]

Question:

Find the constant of proportionality k. Then write an equation for the relationship between x and y

$\begin{array}{ccccc}x & {2} & {4} & {6} & {8} \ \\ y & {10} & {20} & {30} & {40} \ \ \end{array}$

Answer:

(a) $k = 5$

(b) $y = 5x$

Step-by-step explanation:

Given

$\begin{array}{ccccc}x & {2} & {4} & {6} & {8} \ \\ y & {10} & {20} & {30} & {40} \ \ \end{array}$

Solving (a): The constant of proportionality:

Pick any two corresponding x and y values

$(x_1,y_1) = (2,10)$

$(x_2,y_2) = (6,30)$

The constant of proportionality k is:

$k = \frac{y_2 - y_1}{x_2 - x_1}$

$k = \frac{30-10}{6-2}$

$k = \frac{20}{4}$

$k = 5$

Solving (b): The equation

In (a), we have:

$(x_1,y_1) = (2,10)$

k can also be expressed as:

$k = \frac{y- y_1}{x- x_1}$

Substitute values for x1, y1 and k

$5 = \frac{y- 10}{x- 2}$

Cross multiply:

$y - 10 = 5(x - 2)$

Open bracket

$y - 10 = 5x - 10$

Add 10 to both sides

$y - 10 +10= 5x - 10+10$

$y = 5x$

6 0

3 years ago

What is a, b, and c in this quadratic equation?

Vilka [71]

A = 7
b = -18
c = -52
Do you know what the Quadratic formula is to solve?

4 0

3 years ago

Please help me with my homework

Alona [7]

so for A it is 16 times as large because the area is 3*3 which is 9 but since it is a triangle the area is halved and is 4.5 and divide 72 by 4.5 which gives us 16

then for b I believe that the scale factor is 4

and for C it is asking for the bottom side of the scaled copy not of triangle A and 4*3=12 so the bottom is 12

ok now its complete =)

4 0

3 years ago