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3 years ago

F(x) = 9-3x g(x) = 5x-7 Find f(x)+g(x).

Mathematics

1 answer:

Bezzdna [24]3 years ago

8 0

Answer:

In the problem, the sum of the two functions is 2x + 2

Step-by-step explanation:

For this problem, we have to add together f(x) and g(x).

f(x) = 9 - 3x

g(x) = 5x - 7

(f + g)(x) = (9 - 3x) + (5x - 7)

Combine like terms.

(f + g)(x) = 2x + 2

So, when you combine the two functions together, you will get 2x + 2.

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0.6x-0.1=0.5x+2 The solution set is {_} PLEASE PLEASE PLEASE EXPLAIN HOW

timurjin [86]

Answer:

Step-by-step explanation:

Step 1:

0.6x - 0.1 = 0.5x + 2 Equation

Step 2:

0.1x - 0.1 = 2 Subtract 0.5x on both sides

Step 3:

0.1x = 2.1 Add 0.1 on both sides

Step 4:

x = 2.1 ÷ 0.1 Divide

Ansewr:

x = 21

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inn [45]

Answer:

$y=2\sqrt{10}$

Step-by-step explanation:

By applying geometric mean theorem in the given right angle,

$\frac{AD}{DB}=\frac{DB}{CD}$

By substituting values in the given ratio,

$\frac{5}{y}= \frac{y}{8}$

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$y^2=40$

$y=2\sqrt{10}$

Exact value of y is $2\sqrt{10}$ .

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What basic trigonometric identity would you use to verify that

Dafna1 [17]

Given the equation:

$\frac{\sin^2x+\text{cos}^2x}{\cos x}=\sec x$

Let's determine the trigonometric identity that you could be used to verify the exquation.

Let's determine the identity:

Apply the trigonometric identity:

$\sin ^2x+\cos ^2x=1$

$\cos x=\frac{1}{\sec x}$

Replace cosx for 1/secx

Thus, we have:

$\begin{gathered} \frac{\sin^2x+\cos^2x}{\frac{1}{\sec x}} \\ \\ =(\sin ^2x+\cos ^2x)(\sec x) \\ \text{Where:} \\ (\sin ^2x+\cos ^2x)=1 \\ \\ We\text{ have:} \\ (\sin ^2x+\cos ^2x)(\sec x)=1\sec x=\sec x \end{gathered}$

The equation is an identity.

Therefore, the trignonometric identity you would use to verify the equation is:

$\cos ^2x+\sin ^2x=1$

ANSWER:

$\cos ^2x+\sin ^2x=1$

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1 year ago