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3 years ago

Solve the system of equations using >>the Substitution Method<<

1 answer:

6 0

HELLO...!!!

$\begin{cases}&y=2x+4\\&3x+y=9\end{cases}$

$\underline{\textbf{Add\ "y"\ in\ Eq.\ 2:}}\\ \\3x+2x+4=9\\ \\5x+4=9\\ \\5x=9-4\\ \\5x=5\\ \\x= \dfrac{5}{5}\\ \\x=1\quad\checkmark\\ \\ \\\underline{\textbf{Add\ "x"\ in\ Eq.\ 1:}}\\ \\y=2(1)+4\\ \\y=2+4\\ \\y=6\quad\checkmark\\ \\ \\\mathbb{ANSWER:}\Longrightarrow\boxed{\boxed{\boldsymbol{x=1\ ;\ y=6}}}\\ \\ \\\textbf{HOPE\ THIS\ HELPS...!!}$

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The circumference of a circle with a diameter 1.8cm take pie to be 3.142

Gennadij [26K]

The formula to get the circumference is 2πr, π being pi and r being your radius.

Your r is 0.9 since the radius is half the diameter.

2πr = 2 x 3.142 x 0.9

= 5.6556cm

≈ 5.7cm

4 0

2 years ago

What is x if 3x-2=x-3

blsea [12.9K]

Answer:

x=-1/2

Step-by-step explanation:

add 2 to both sides

3x=x-1

subtract x from both sides

2x=-1

divide by 2

x=-1/2

7 0

3 years ago

Read 2 more answers

Would you use SSS or SAS to prove the triangles congruent? If there is not enough information to prove by SSS or SAS, then write

egoroff_w [7]

<h2><u>Answer (d</u>) :</h2>

These triangles are congruent under the SAS congruence criterion. We can prove it using the following steps :

Given :

In triangle 1 and 2 :

A side in triangle 1 and a side in triangle 2 are equal.
Another side in triangle 1 and triangle 2 is equal.
The angle between the equivalent sides is also equal.

Since the SAS congruence criterion says that two triangles are congruent if two of their sides and one included angle is equal and since these 2 triangles fulfill all of those rules, we can conclude that these triangles are congruent under the SAS congruence criterion.

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2 years ago

Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One common ch

enot [183]

Answer:

$y=\frac{-7t^2+22t-7}{7t-22}$

Step-by-step explanation:

We are given that

Initial value problem

$y'=(t+y)^2-1$ , y(3)=4

Substitute the value $z=t+y$

When t=3 and y=4 then

z=3+4=7

$y'=z^2-1$

Differentiate z w.r.t t

Then, we get

$\frac{dz}{dt}=1+y'$

$z'=1+z^2-1=z^2$

$z^{-2}dz=dt$

Integrate on both sides

$-\frac{1}{z}dz=t+C$

$z=-\frac{1}{t+C}$

Substitute t=3 and z=7

Then, we get

$7=-\frac{1}{3+C}$

$21+7C=-1$

$7C=-1-21=-22$

$C=-\frac{22}{7}$

Substitute the value of C then we get

$z=-\frac{1}{t-\frac{22}{7}}$

$z=\frac{-7}{7t-22}$

$y=z-t$

$y=\frac{-7}{7t-22}-t$

$y=\frac{-7-7t^2+22t}{7t-22}$

$y=\frac{-7t^2+22t-7}{7t-22}$

8 0

3 years ago

What does it mean when a equation has no solution?

velikii [3]

Hello!

No solutions means that there is two sides that do not equal or math, therefore, your answer would be c. both sides of the equation are not equal to one another.
Enjoy.

Brainliest please?

4 0

3 years ago

Read 2 more answers