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2 years ago

8

Which statements are true about tessellations?

1 answer:

PIT_PIT [208]2 years ago

6 0

The correct answer is 1,3& 4.

A figure can tessellate only if its interior angle is a factor of 360. For an equilateral triangle, each interior angle is 60°; 360÷60 = 6, so this works.

For a regular hexagon, each interior angle is (6-2)(180)/6=4(180)/6=720/6=120; 360÷120=3, so this works.

For a regular pentagon, each interior angle is (5-2)(180)/5=3(180)/5=540/5=108; 360÷108=3.3. This does not divide evenly, so it does not work.

For a square, each interior angle is 90°; 360÷90=4, so this works.

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Is the equation below written in standard form? If not, select which explanation best applies to why the equation

natita [175]

Answer:

D

Step-by-step explanation:

The equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers

12 = 2x + 4y , that is

2x + 4y = 12 ← is in standard form

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2 years ago

Tell whether each proportion is true or false and prove your answer.

Arte-miy333 [17]

Answer:

the proportion is false

Step-by-step explanation:

5.5/7.5 = 11/15

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2 years ago

ANSWER ASAP PLEASE

inna [77]

Same as the dude that did the same thing

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2 years ago

Read 2 more answers

44.1 = 4.9p what does “p” equal

Talja [164]

Step 1: Flip the equation.

4.9p=44.1

Step 2: Divide both sides by 4.9.

4.9p= 44.1

4.9/4.9 = 44.1/4.9

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2 years ago

Solve the following equations.<br> log2(x^2 − 16) − log^2(x − 4) = 1

Alenkasestr [34]

Answer:

$x=\frac{4*(2+e)}{e-2}$

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

$log(\frac{1}{x} )=-log(x)$

$log(x*y)=log(x)+log(y)$

Therefore:

$log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1$

Now, cancel logarithms by taking exp of both sides:

$e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e$

Multiply both sides by $(x-4)^{2}$ and using distributive propierty:

$2x^{2} -32=16e-8ex+ex^{2}$

Substract $16e-8ex+ex^{2}$ from both sides and factoring:

$-(x-4)*(-8-4e-2x+ex)=0$

Multiply both sides by -1:

$(x-4)*(-8-4e-2x+ex)=0$

Split into two equations:

$x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0$

Solving for $x-4=0$

Add 4 to both sides:

$x=4$

Solving for $-8-4e-2x+ex=0$

Collect in terms of x and add $4e+8$ to both sides:

$x(e-2)=4e+8$

Divide both sides by e-2:

$x=\frac{4*(2+e)}{e-2}$

The solutions are:

$x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}$

If we evaluate x=4 in the original equation:

$log(0)-log(0)=1$

This is an absurd because log (x) is undefined for $x\leq 0$

If we evaluate $x=\frac{4*(2+e)}{e-2}$ in the original equation:

$log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1$

Which is correct, therefore the solution is:

$x=\frac{4*(2+e)}{e-2}$

6 0

3 years ago