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2 years ago

9

I need help on my math qustion 6.02 i'm in k12

1 answer:

Nana76 [90]2 years ago

3 0

This is how it should look
9 15 24
12 10 22
21 25 46

I hope this helps and this is how it has to look just like this

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olya-2409 [2.1K]

Here's the solutions :

1. solve for a :

$4a = 20$

$a = \dfrac{20}{4}$

$a = 5$

2. solve for y :

$3y + 9 = 24$

$3y = 24 - 9$

$3y = 15$

$y = \dfrac{15}{3}$

$y = 5$

3. solve for y :

$\dfrac{y}{3} - 5 = 4$

$\dfrac{y}{3} = 5 + 4$

$\dfrac{y}{3} = 9$

$y = 9 \times 3$

$y = 27$

4. solve for k :

$3 = 9 - 4k$

$3 - 9 = -4 k$

$- 6 = - 4 k$

$k = \dfrac{ - 6}{ - 4}$

$k = \dfrac{3}{2} \: \: \: or \: \: 1.5$

5. there's something wrong with this question.

6. solve for x :

$3x + 8 = 2$

$3x = 2 - 8$

$3x = - 6$

$x = \dfrac{ - 6}{3}$

$x = - 2$

i hope it helped...

6 0

2 years ago

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Tính tích phân sau bằng cách dùng tọa độ cực I=∫∫ <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%5E%7B2%7D%20%2B

xenn [34]

It sounds like <em>R</em> is the region (in polar coordinates)

<em>R</em> = {(<em>r</em>, <em>θ</em>) : 2 ≤ <em>r</em> ≤ 3 and 0 ≤ <em>θ</em> ≤ <em>π</em>/2}

Then the integral is

$\displaystyle \iint_R\frac{\mathrm dx\,\mathrm dy}{\sqrt{x^2+y^2}} = \int_0^{\pi/2}\int_2^3 \frac{r\,\mathrm dr\,\mathrm d\theta}{\sqrt{r^2}} \\\\ = \int_0^{\pi/2}\int_2^3 \mathrm dr\,\mathrm d\theta \\\\ = \frac\pi2\int_2^3 \mathrm dr \\\\ = \frac\pi2r\bigg|_2^3 = \frac\pi2 (3-2) = \boxed{\frac\pi2}$

5 0

2 years ago

Help me out please?<br>chapter : exponents and powers

belka [17]

Answer:

$\frac{125}{27}$

Step-by-step explanation:

$\frac{2^{-4} \times 15^{-3} \times 625}{5^2 \times 10^{-4}}$

Lets expand all the composite numbers into prime numbers.

$=> \frac{2^{-4} \times (3^{-3} \times 5^{-3}) \times 5^4}{5^2 \times (2^{-4} \times 5^{-4})}$

Lets cancel $2^{-4}$ from numerator and denominator.

$=> \frac{3^{-3} \times 5^{-3} \times 5^4}{5^2 \times 5^{-4}}$

Using laws of exponents , lets solve this.

$=> \frac{3^{-3} \times 5^{(-3 + 4)}}{5^{( 2 - 4)}}$

$=> \frac{3^{-3} \times 5^{1}}{5^{-2}}$

$=> 3^{-3} \times 5^{[1 - (-2)]}$

$=> 3^{-3} \times 5^{3}$

$=> \frac{5^3}{3^3} = \frac{125}{27}$

4 0

2 years ago

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Need some help with this question

adelina 88 [10]

Answer:

iba nalang

Step-by-step explanation:

just tell someting wag yan

6 0

2 years ago

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4) If g(x) = 13x2 what is the value of g(-3)?

juin [17]

(6) typing for 20 characters dah dah dah

4 0

2 years ago

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