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4 years ago

Best describes the expression 7(y+5)?

Mathematics

1 answer:

stiv31 [10]4 years ago

6 0

Answer:

Step-by-step explanation:

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Someone help me asap

leva [86]

Answer:

The correct answer is – 1/2

6 0

3 years ago

Read 2 more answers

Solve each system using a matrix

Serjik [45]

<u>ANSWER </u>

$x=\frac{1}{2}$ and $y=\frac{1}{4}$

<u>EXPLANATION</u>

Given;

$4x-12y=1$

and

$6x+4y=4$

The augmented matrix of the two linear equation is given by;

$\left[\begin{array}{ccc}4&-12|&-1\\6\:&\:\:\:4|&4\end{array}\right]$

We now perform row operations;

$\frac{1}{4}\times R_1 \rightarrow R_1$ .

This gives us

$\left[\begin{array}{ccc}1&-3|&\frac{-1}{4}\\6\:&\:\:\:4|&4\end{array}\right]$

$R_2-6R_1 \rightarrow R_2$

$\left[\begin{array}{ccc}1&-3|&\frac{-1}{4}\\0\:&\:\:\:22|&\frac{11}{2}\end{array}\right]$

$\frac{1}{22}R _2 \rightarrow R_2$

$\left[\begin{array}{ccc}1&-3|&\frac{-1}{4}\\0\:&\:\:\:1|&\frac{1}{4}\end{array}\right]$

$R_1+3R_2 \rightarrow R_1$

$\left[\begin{array}{ccc}1&0|&\frac{1}{2}\\0\:&\:\:\:1|&\frac{1}{4}\end{array}\right]$

Hence $x=\frac{1}{2}$ and $y=\frac{1}{4}$

4 0

3 years ago

Find the value of x.

nlexa [21]

A triangle =180
the box in the corner =90°
so, 180-90-20 =70

x=70

7 0

3 years ago

GEOMETRY HELP AHH please

KATRIN_1 [288]

Answer:

Step-by-step explanation:

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3 years ago

Solve the differential equation y" + 3y' + 2y = P1 where, y" = d2y/dx2, y' = dy/dx, P1 = P1(x)) and the initial conditions are y

Ne4ueva [31]

Answer:

$y(x) = 0$

Step-by-step explanation:

Let P(1) = 0, the differential equation becomes:

y" + 3y' + 2y = 0

Also let y' = my and y'' = m²y

Substitute

m²y+3my+2y = 0

(m²+3m+2)y = 0

The auxiliary equation is expressed as:

m²+3m+2 = 0

Factorize

m²+2m+m+2 = 0

m(m+2)+1(m+2) = 0

(m+1)(m+2) = 0

m+1 = 0 and m+2 = 0

m₁ = -1 and m₂ = -2

Since the value of m is real and distinct, the solution to the differential equation will be expressed as:

$y = C_1e^{m_1x}+C_2e^{m_2x}\\$

$y(x) = C_1e^{-x}+C_2e^{-2x}\\$

Given the initial condition y(0) = 0

$y(0) = C_1e^{-0}+C_2e^{-2(0)}\\0 = C_1+C_2\\C_1+C_2 = 0 ................. 1$

If y'(0) = 0

$y'(x) = -C_1e^{-x}-2C_2e^{-2x}\\y'(0) = -C_1e^{-0}-2C_2e^{-2(0)}\\0 = -C_1-2C_2\\C_1-2C_2 = 0...................... 2$

Solve eqn 1 and 2 simultaneously

From 1:

C₁ = -C₂ .......... 3

Substitute equation 3 into 2:

-C₂-2C₂ = 0

-3C₂ = 0

C₂ = 0

Substitute the constant into the differential equation.

$y(x) = C_1e^{-x}+C_2e^{-2x}\\\\y(x) = 0$

7 0

4 years ago