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3 years ago

9

What is an scale and interval

1 answer:

den301095 [7]3 years ago

4 0

Answer:

The space between each value on the scale of a bar graph is called an interval. In other words, the interval is the relation between the units you're using, and their representation on the graph, or the distance between marks. You choose intervals based on the range of the values in the data set.^-^

Step-by-step explanation:

Hoped I Have Helped Have Nice Day "Peace"

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Determine the number of real solutions the equation has.

just olya [345]

Using the quadratic formula, the solutions are x1 = 1.3416 and x2 = -1.3416
So, it has 2 real solutions.

7 0

3 years ago

Solve this equation. <br> Leave the answer as an improper fraction. <br> 3x^2 + 2 = 11 - x^2

Nikolay [14]

Answer:

$x = \frac{3}{2}orx=-\frac{3}{2}$

Step-by-step explanation:

The given Equation is

$3x^{2} + 2 = 11-x^{2}$ .................(i)

Adding ( $x^{2} -11$ ) on both sides of (i)

$3x^{2}+2+x^{2}-11 = 11-x^{2}+x^{2}-11$

solving and cancelling out the terms will give

⇒ $4x^{2} -9=0$

⇒ $(2x)^{2} -(3)^{2} = 0$ ...................(ii)

Now we Know that

$(a)^{2} -(b)^{2} = (a-b)(a+b)$

Applying this on equation (ii)

$(2x-3)(2x+3)=0$

This will lead us to

Either 2x-3 =0 .......(iii) or 2x+3=0 ..........(iv)

Solving equation (iii) for value of x

2x - 3 =0

adding 3 on both sides

2x -3 +3 = 0+ 3

2x = 3

Cross multiplying gives

$x=\frac{3}{2}$

Solving equation (iv) for value of x

2x + 3 =0

adding -3 on both sides

2x -3 +3 = 0- 3

2x = -3

Cross multiplying gives

$x=\frac{-3}{2}$

so $x=\frac{3}{2}$ and $x=\frac{-3}{2}$

7 0

3 years ago

2/3x=10 thankyou for helping me

Novay_Z [31]

(2/3)x = 10
(3/2)(2/3)x = 10(3/2)
x = 30/2 = 15

6 0

3 years ago

Read 2 more answers

A parcel delivery service will deliver a package only if the length plus girth (distance around) does not exceed 84 inches. (A

Kisachek [45]

Answer:

A. 14x14x28

B. The maximum volume is 5488 cuibic inches

Step-by-step explanation:

The problem states that the box has square ends, so you can express volume with:

$v=x^{2} y$

Using the restriction stated in the problem to get another equation you can substitute in the one above:

$4x+y=84\\\\$

Substituting <em>y</em> whit this equation gives:

$v=x^{2} (84-4x)\\\\v=84x^{2} -4x^{3}$

Now find the limit of <em>x</em>:

$\frac{84x^{2}-4x^{3}}{dx}=168x-12x^{2}\\\\x=\frac{168}{12}=14$

Find the length:

$y=84-4(14)=28$

You can now calculate the maximum volume:

$v=(14)^{2}(28)= 5488$

6 0

3 years ago

What’s the distance between points (5,6) and (9,9)

iragen [17]

Answer:

(4,3)

the distance between x1 and x2 then y1 then y2

6 0

3 years ago