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3 years ago

12

it's sixth grade math I'm sorry for the words that are cut off you'll be able to figure out what they are i dont have a better p

hoto

1 answer:

alekssr [168]3 years ago

7 0

8x6=48 min

Hope I helped!

You might be interested in

when 6 is subtracted from the square of a number, the result is 5 times the number. Find the negative solution.

sveta [45]

When 6 is subtracted from the square of a number, the result is 5 times the number, then the negative solution is -1

<h3><u>Solution:</u></h3>

Given that when 6 is subtracted from the square of a number, the result is 5 times the number

To find: negative solution

Let "a" be the unknown number

Let us analyse the given sentence

square of a number = $a^2$

6 is subtracted from the square of a number = $a^2 - 6$

5 times the number = $5 \times a$

<em><u>So we can frame a equation as:</u></em>

6 is subtracted from the square of a number = 5 times the number

$a^2 - 6 = 5 \times a\\\\a^2 -6 -5a = 0\\\\a^2 -5a -6 = 0$

<em><u>Let us solve the above quadratic equation</u></em>

For a quadratic equation $ax^2 + bx + c = 0$ where $a \neq 0$

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here in this problem,

$a^2-5 a-6=0 \text { we have } a=1 \text { and } b=-5 \text { and } c=-6$

Substituting the values in above quadratic formula, we get

$\begin{array}{l}{a=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(-6)}}{2 \times 1}} \\\\ {a=\frac{5 \pm \sqrt{25+16}}{2}=\frac{5 \pm \sqrt{49}}{2}} \\\\ {a=\frac{5 \pm 7}{2}}\end{array}$

We have two solutions for "a"

$\begin{array}{l}{a=\frac{5+7}{2} \text { and } a=\frac{5-7}{2}} \\\\ {a=\frac{12}{2} \text { and } a=\frac{-2}{2}}\end{array}$

<h3>a = 6 or a = -1</h3>

We have asked negative solution. So a = -1

Thus the negative solution is -1

6 0

3 years ago

10. When you add the same number to both sides of an inequality, is the inequality still true? Explain how you know your stateme

avanturin [10]

The inequality is still true! If you add a number, say 5 to both sides of the following inequality, does anything change?

3 < 6

3 + 5 < 6 + 5

8 < 11

The inequality is still true. We know the statement holds for subtracting the same number because, in a way, addition and subtraction are pretty much the same operation. If I subtract 5 from both sides, I can think of it like "I add negative 5 to both sides" or something along those lines. It's kind of backwards thinking.

6 0

4 years ago

Is y=5x-25 proportional

Vinvika [58]

It is proportional because we increase y.

3 0

4 years ago

dybincka [34]

Answer:

Absolute minimum = 1.414

Absolute maximum = 2.828

Step-by-step explanation:

$g(x,y)=\sqrt {x^2+y^2} \ constraints: 1\leq x\leq 2 ,\ 1\leq y\leq2$

For absolute minimum we take the minimum values of $x$ and $y$ .

$x_{minimum} =1\\y_{minimum}=1\\$

Plugging in the minimum values in the function.

$g(1,1)=\sqrt {1^2+1^2}\\g(1,1) = \sqrt{1+1}\\g(1,1)=\sqrt {2}\\g(1,1)=\pm 1.414\\$

Absolute minimum value will be always positive.

∴ Absolute minimum = 1.414

For absolute maximum we take the maximum values of $x$ and $y$ .

$x_{maximum} =2\\y_{maximum}=2\\$

Plugging in the maximum values in the function.

$g(2,2)=\sqrt {2^2+2^2}\\g(2,2) = \sqrt{4+4}\\g(2,2)=\sqrt {8}\\g(2,2)=\pm 2.828\\$

Absolute maximum value will be always positive.

∴ Absolute maximum = 2.828

3 0

4 years ago

DOES ANYONE UNDERTAND THIS? IF SO PLEASE HELP

Nana76 [90]

The sides must be equal otherwise BC and AD wouldn't be parallel. So
12x-6 = 2x+7
10x = 13
x = 13/10 = 1.3

8 0

4 years ago