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3 years ago

10

Larry used a pattern of colors to make a cube train. He used a red cube, a blue cube, a red cube, and another red cube before he

started the pattern again. How many red cube did Larry use?

1 answer:

OlgaM077 [116]3 years ago

8 0

I think 18 red cubes.
Explanation: A cube has 6 sides, so you would use the pattern 6 times. There are 3 red cubes in each pattern, you use the pattern 6 times, meaning 3 × 6, so the cube train has 18 red cubes. I hope this helps!

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What number has the same value as 14 tens

nasty-shy [4]

The number with the same value will be 40 because 10 tens make 100 and 4 tens make 40 so it wili be 140

8 0

3 years ago

Read 2 more answers

Find the value of c that makes the expression a perfect square trinomial.<br><br> x2 + 4x + c

masha68 [24]

the value of c that makes the expression a perfect square binomial is c=4 .

<u>Step-by-step explanation:</u>

Here we have , an expression x2 + 4x + c or , $x^2 + 4x + c$ . We need to find the value of c that makes the expression a perfect square binomial. Let's find out:

We have , $x^2 + 4x + c$

⇒ $x^2+4x+c$

⇒ $x^2+2(2)x+c$

Now , we know that $(a+b)^2=a^2+2(a)(b)+b^2$

Comparing above equation , to $x^2+2(2)x+c$ we get ;

⇒ $x^2+2(2)x+c$ { $c=2^2$ }

⇒ $x^2+2(2)x+2^2$

⇒ $(x+2)^2$

Therefore , the value of c that makes the expression a perfect square binomial is c=4 .

7 0

3 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

4 years ago

Is a triangle with side lengths 8, 12, and 16 a right triangle

Nataly_w [17]

Answer:

No.

Step-by-step explanation:

5 0

3 years ago

Which equations are true for x = –2 and x = 2? Select two options

yulyashka [42]

The equations that are true for x = -2 and x = 2 are $x^{2}$ -4 = 0 and 4 $x^{2}$ = 16

<h3>What is an equation?</h3>

An equation is a mathematical statement that establishes the equality of two quantities.

Analysis:

Testing the equation x2 - 4 = 0

for x = 2, $(2)^{2}$ - 4 = 4-4 = 0

for x = -2 = (-2)2 - 4 = 4-4 = 0

Therefore x2-4 = 0 is true

Testing for 4x2 =16

for x = 2 , 4(2) ^2 = 4x4 = 16

for x = -2 4(-2)^2 = 4(4) = 16.

All other equations are not true except these two.

Learn more about equations: brainly.com/question/2972832

#SPJ1

6 0

2 years ago