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3 years ago

12

Chloe is sorting items at a thrift store. She finds a box if tiny blocks that has 1,344 written on the outside. she can see that

each layer in the box has 64 blocks.

1 answer:

ozzi3 years ago

5 0

21 layers of blocks if that's what you're looking for

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Isaac invested $1,800 in an account paying an interest rate of 4.7%

bazaltina [42]

Answer:

$3584.86 or 3600 if rounded to the nearest 100 dollars

Step-by-step explanation:

1800(1.047)^15= 3584.86

you can plug this into a calculator

7 0

3 years ago

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What is the next term in the pattern shown below? 10, 8.9, 7.8, 6.7, ...A. 6.6 C. 5.5B. 5.6 D. 1.1 please help!!

anzhelika [568]

You subtract 1.1 each time so the answer is 6.6 (a)

6 0

3 years ago

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Pls answer and let me know if i can factor this r not

enyata [817]

Answer:

18r-4

Step-by-step explanation:

Here is your anwer

6 0

3 years ago

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If 2^(x+3) - 2x = k(2^x) what is the value of k

svlad2 [7]

Answer:

$\large\boxed{C)\ 7}$

Step-by-step explanation:

Use:

$a^n\cdot a^m=a^{n+m}$

$2^{x+3}-2^x=2^x\cdot2^3-2^x=8\cdot2^x-2^x=(8-1)\cdot2^x=7\cdot2^x\\\\2^{x+3}-2^x=k(2^x)\\\\7\cdot2^x=k(2^x)\Rightarrow k=7$

$8\cdot2^x-2^x=8\cdot2^x-1\cdot2^x=(8-1)(2^x)=7\cdot2^x\\\\\text{distributive property:}\ a(b-c)=ab-ac$

8 0

4 years ago

If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?

lys-0071 [83]

Answer:

2

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

$3x^2-(3k-2)x-(k-6) \text{ with }k=2$ :

$3x^2-(3\cdot 2-2)x-(2-6)$

$3x^2-4x+4$

I'm going to solve $3x^2-4x+4=0$ for x using the quadratic formula:

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}$

$\frac{4\pm \sqrt{16-16(3)}}{6}$

$\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}$

$\frac{4\pm 4\sqrt{-2}}{6}$

$\frac{2\pm 2\sqrt{-2}}{3}$

$\frac{2\pm 2i\sqrt{2}}{3}$

Let's see if uv=u+v holds.

$uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}$

Keep in mind you are multiplying conjugates:

$uv=\frac{1}{9}(4-4i^2(2))$

$uv=\frac{1}{9}(4+4(2))$

$uv=\frac{12}{9}=\frac{4}{3}$

Let's see what u+v is now:

$u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}$

$u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$

We have confirmed uv=u+v for k=2.

4 0

3 years ago