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3 years ago

2. Solve the system using substitution. (1 point)

Mathematics

2 answers:

AveGali [126]3 years ago

7 0

Answer:

(8, 11 )

Step-by-step explanation:

Given the 2 equations

2x + 2y = 38 → (1)

y = x + 3 → (2)

Substitute y = x + 3 into (1)

2x + 2(x + 3) = 38 ← distribute and simplify left side

2x + 2x + 6 = 38

4x + 6 = 38 ( subtract 6 from both sides )

4x = 32 ( divide both sides by 4 )

x = 8

Substitute x = 8 into (2) for corresponding value of y

y = 8 + 3 = 11

Solution is (8, 11 )

Pepsi [2]3 years ago

7 0

It’s (8,11) if you sub everything

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What is f(x)=2(3)^x?

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Answer:

This is an exponential function

Step-by-step explanation:

f(x)=2(3)^x is one example of the exponential function. Since f(0) = 2(1) = 2, the y-intercept is (0, 2). The x-axis is the horizontal asymptote. The graph begins in Quadrant II, passes through (0, 2) and continues to increase, faster and faster, in Quadrant I. Next time, please be more specific about what you'd like to know.

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One student can paint a wall in 16 minutes. Another student can paint the same wall in 24 minutes. Working together, about how l

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Let's look at work rates per minute.

Together they can paint the wall in x minutes.
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3 years ago

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3 years ago

Please help solve for x..

aalyn [17]

Answer:

x = 1 or x = -1

Step-by-step explanation:

Given equation:

$x^{100}-4^x \cdot x^{98}-x^2+4^x=0$

Factor out -1:

$\implies -1(-x^{100}+4^x \cdot x^{98}+x^2-4^x)=0$

Divide both sides by -1:

$\implies -x^{100}+4^x \cdot x^{98}+x^2-4^x=0$

Rearrange the terms:

$\implies 4^x \cdot x^{98}-4^x-x^{100}+x^2=0$

$\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c \quad \textsf{to }x^{100}:$

$\implies 4^x \cdot x^{98}-4^x-x^{98}x^2+x^2=0$

Factor the first two terms and the last two terms separately:

$\implies 4^x(x^{98}-1)-x^2(x^{98}-1)=0$

Factor out the common term $(x^{98}-1)$ :

$\implies (4^x-x^2)(x^{98}-1)=0$

Zero Product Property: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).

Using the Zero Product Property, set each factor equal to zero and solve for x (if possible):

$\begin{aligned}x^{98}-1 & = 0 & \quad \textsf{or} \quad \quad4^x-x^2 & = 0 \\x^{98} & =1 & 4^x & = x^2 \\x & = 1, -1 & \textsf{no}& \textsf{ solutions for } x \in \mathbb{R}\end{aligned}$

Therefore, the solutions to the given equation are: x = 1 or x = -1

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