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blagie [28]
3 years ago
8

Math word problem involving reducing a shape

Mathematics
1 answer:
jekas [21]3 years ago
8 0
John decides to reduce a shape one half of its original size. The square that he had originally is 4 x 4. He reduces it so it will be 2×2
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Someone please I just want to get done with this
marusya05 [52]

Answer:

{10, 20, 30, 40, 50, 60}

Step-by-step explanation:

The function is:

f(x)=10x\\\{x\in \mathbb{N}|1\leq x \leq 6\}

State that the tange of function is the set of all output values of a function. Therefore, the answer cannot be {1, 2, 3, 4, 5, 6} and 1\leq x\leq 6. The answer is not $10\leq y\leq 60$ because its output will always be a natural number and it's not stated in the answer, like I did above. The answer is {10, 20, 30, 40, 50, 60}

4 0
3 years ago
What s the equation for y= (0,2) slope= -3/7
masya89 [10]
The standard form equation for a straight line is y=mx+b where m is the slope and b is the y intercept. Substitute your given information:
y= -3/7x+2 is the equation
5 0
3 years ago
What is number 1 and 2
ioda
yes the number is is proportional . 3$ a rose in sets of 3 and up . that give you a price for 15 at 45 $ so the cost of 30 Rose's would be 90$.
6 0
3 years ago
Which shows a reflection over the x-axis?
a_sh-v [17]

Answer:

Probably a bit late but its A the first one ;P

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Solve the following equation by completing the square. 3x^2-3x-5=13
mr Goodwill [35]

we'll start off by grouping some

\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}

6 0
3 years ago
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