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Nadya [2.5K]
2 years ago
5

Please answer need now

Mathematics
1 answer:
topjm [15]2 years ago
4 0

Explanation:

A. When you examine the values in the table, you see the y-value is not proportional to the x-value. That is, 20/1 ≠ 26/2. However, you do notice that the y-values increase by 6 when the x-values increase by 1. <em>The ratio of these increases is the slope of the graph of the relation between y and x</em>.

Since the slope is 6, we might expect a value of 6×1 = 6 for x=1. However, the y-value for x=1 is 20, which is 14 more than what might be predicted by the slope. <em>This added extra is the y-intercept on the graph of the relation</em>.

The slope is 6 and the y-intercept is 14.

__

B. The slope is (26-20)/(2-1) = 6.

__

C. For x=1, the slope-intercept equation tells us ...

  y = 6x +b

  20 = 6·1 + b . . . . . . . . . . substitute (x, y) = (1, 20)

  20 -6 = 14 = b . . . . . . . . subtract 6 to find b

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PLEASE HELP FAST!!!!!!!!!
tresset_1 [31]
B is definitely the answer
4 0
3 years ago
Pls help me i will mark brainliest if you are right !
tekilochka [14]

Answer:

second option

Step-by-step explanation:

diameter= 22 cm

radius = diameter/2

=22/2

11 cm

volume of a sphere = \frac{4}{3}*pie* r^3

=\frac{4}{3} * 3.14 *11^3

=4.186666 * 1331

=5572 . 5 cubic centimeters (cm^3)

5 0
3 years ago
What’s is the solution to the equation x to the power of 2 =16/25
skelet666 [1.2K]

Answer:

x=\dfrac{4}{5}

Step-by-step explanation:

The given equation is :

x^2=\dfrac{16}{25}

We need to find the value of x.

When the exponent 2 removes from LHS, it will shift to RHS as square root as follows :

x=\sqrt{\dfrac{16}{25}}

We know that, 4×4= 16 and 5×5 = 25

So,

x=\sqrt{\dfrac{4\times 4}{5\times 5}}\\\\x=\dfrac{4}{5}

So, the solution of the given equation is 4/5.

4 0
2 years ago
What are these questions
Anna007 [38]

Answer:


Step-by-step explanation:

Fractions

7 0
3 years ago
What is the following sum? Assume x20 and 20.<br> √√x²y³ +2√√x³y² + xy √√y
max2010maxim [7]

The value of the expression \sqrt{x^2y^3} + 2\sqrt{x^3y^4} + xy\sqrt y is 2xy\sqrt{y} + 2xy^2\sqrt{x}

<h3>How to evaluate the sum?</h3>

The attachment represents the proper format of the question

The summation expression is given as:

\sqrt{x^2y^3} + 2\sqrt{x^3y^4} + xy\sqrt y

Expand the radicands

\sqrt{x^2y^2 * y} + 2\sqrt{x^2y^4* x} + xy\sqrt y

Evaluate the square roots

xy\sqrt{y} + 2xy^2\sqrt{x} + xy\sqrt y

Add the like terms

2xy\sqrt{y} + 2xy^2\sqrt{x}

Hence, the value of the expression \sqrt{x^2y^3} + 2\sqrt{x^3y^4} + xy\sqrt y is 2xy\sqrt{y} + 2xy^2\sqrt{x}

Read more about expressions at:

brainly.com/question/723406

#SPJ1

3 0
2 years ago
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