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2 years ago

8 x 10*-3

Mathematics

1 answer:

NARA [144]2 years ago

7 0

Answer:

$\huge\boxed{2\times10^3}$

Step-by-step explanation:

$(8\times10^{-3})\div(4\times10^{-6})=\dfrac{8\times10^{-3}}{4\times10^{-6}}=\dfrac{8}{4}\times10^{-3-(-6)}=2\times10^{-3+6}=2\times10^3$

Used:

$\dfrac{a^n}{a^m}=a^{n-m}\ \text{for}\ a\neq0$

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Please solve this by solving systems by elimination 6x+3y=-18 7x+7y=0

gavmur [86]

Answer:

Step-by-step explanation:

6x + 3y = -18

Taking 3 common from left side

so, 3(2x+y) = -18

2x+y = -18/3

2x+y = -6 (equation 1)

and, 7x + 7y = 0

Taking 7 as commom from left side

7(x + y) = 0

x + y = 0/7

x + y = 0 (equation 2)

now , by using elimination method

subtracting equation2 from equation1

2x + y - (x + y)= -6 - 0

2x + y - x - y = -6

x = -6

so, x = -6

substituting value of x in equation 1

2(-6) + y = -6

-12 + y = -6

y = -6 +12

y = 6

hence, the value of x = -6

and value of y = 6

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3 years ago

If a pen and a pencil cost 1.10$ but the pen cost is 1$ more than pencil

Stella [2.4K]

Answer:

The pencil cost $0.05 and the pen cost $1.05.

Step-by-step explanation:

Let, the price of the pencil is $ .

Given that the pen costs a dollar more than the pencil,

So, the price of the pen=$+1.

Therefore, ++1=1.10

, 2+1=1.10

, 2=1.10–1

, 2=0.10

, =0.10/2

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The pencil cost $0.05 and the pen cost $1.05.

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2 years ago

Find the slope intercept form for the line satisfying the following conditions. Passing through (3,-5) and (5,-9)

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Answer: m = -4 / 2 = -2 / 1 = -2

Slope (m) = -2

Step-by-step explanation:

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2 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:

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.

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2 years ago

Lee is selling spirit wear for a fundraiser hats cost $8 and shirts $12 lee sold $272 worth of spirit wear in the first month. I

ivolga24 [154]

16*8=128
12*12=144

128
+ 144
$272

Lee sold 16 hats

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3 years ago