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Sergeu [11.5K]
3 years ago
11

Find the value of 6+x when x = 15.

Mathematics
2 answers:
zzz [600]3 years ago
8 0

Answer:

Step-by-step explanation:

Given the function f(x)=6+x

F(x) is dependent on x,

When x=1

f(x)=6+x

f(x)=6+1,

f(x)=7

When x=2

f(x)=6+x

f(x)=6+2

f(x)=8.

This will continue like this till we get to x=15

So when x=15

We will substitute x=15 into the function f(x)

f(x)=6+x

f(x)=6+15

f(x)=21

Then, the answer is 21.

xxMikexx [17]3 years ago
5 0

Answer: 17

Step-by-step explanation:

To solve 6+x when x = 15,

Step 1: Substitute 15 into x

Step 2: Sum 6 and 15

6+15= 17.

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What are the exact solutions of x2 − 5x − 7 = 0, where x equals negative b plus or minus the square root of b squared minus 4 ti
Marat540 [252]

Answer:

x = \frac{5 + \sqrt{53}}{2}  or  x = \frac{5 - \sqrt{53}}{2}

Step-by-step explanation:

Given

x^2 - 5x - 7 = 0

Required

Solve for x using:

x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}

First, we need to identify a, b and c

The general form of a quadratic equation is:

ax^2 + bx + c = 0

So, by comparison with x^2 - 5x - 7 = 0

a = 1     b = -5      c = -7

Substitute these values of a, b and c in

x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}

x = \frac{-(-5) \± \sqrt{(-5)^2 - 4 * 1 * -7}}{2 * 1}

x = \frac{5 \± \sqrt{25 +28}}{2}

x = \frac{5 \± \sqrt{53}}{2}

Split the expression to two

x = \frac{5 + \sqrt{53}}{2}  or  x = \frac{5 - \sqrt{53}}{2}

To solve further in decimal form, we have

x = \frac{5 + 7.28}{2}  or  x = \frac{5 - 7.28}{2}

x = \frac{12.28}{2}  or  x = \frac{-2.28}{2}

x = 6.14 or x = -1.14

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3 years ago
What is greater 2yards or5feet
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2 years ago
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The equation of a circle is (x - 4)2 + (y - 4)2 = 9. The point (4, 1) is on the circle.
Alika [10]

Answer:

4

Step-by-step explanation:

the answer is 4 hope it helps

3 0
1 year ago
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