Ask question

Ask question

All categories

2 years ago

6

find two consecutive odd numbers such that the sum of the larger number and twice the smaller number is 27 less than four times

the smaller number

1 answer:

Vlad1618 [11]2 years ago

7 0

Answer:

DANG some of these questions are from like 2017 and still havent been answerd

Step-by-step explanation:

You might be interested in

Margo soccer coach brings a large bag of watermelon to practice.the mass of the bag with all the watermelon in it is 12 kg. How

nirvana33 [79]

6 watermelons, because each watermelon is 2 kg

4 0

2 years ago

Jenny's chocolate bar is 34% cocoa. If the weight of the chocolate bar is 66 grams, how many grams of cocoa does it contain? Rou

damaskus [11]

Answer:

22.4 grams

Step-by-step explanation:

66 times 0.34 is 22.44

22.44 rounded to the nearest tenth is 22.4

hope i helped :)

7 0

2 years ago

Read 2 more answers

Hi there can anyone help me with this please

kipiarov [429]

I think it is 64. you add all of them together and hten divide it by 6.

3 0

3 years ago

How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?

MrRissso [65]

$\left( \frac{27x^3}{8y^9}\right)^ \frac{5}{3} \\\\\\ =\left( \frac{(3x)^3}{(2y^3)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(3x)^{3 \times \frac{5}{3} }}{(2y^3)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(3x)^5}{(2y^3)^{5 }} \\\\\\ =\frac{243x^5}{32y^{15}}$

Now, If the exponent was negative like you asked....

$\left( \frac{27x^3}{8y^9}\right)^ {-\frac{5}{3}} \\\\\\ =\left( \frac{8y^9}{27x^3}\right)^ {\frac{5}{3}}\\\\\\ =\left( \frac{(2y^3)^3}{(3x)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(2y^3)^{3 \times \frac{5}{3} }}{(3x)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(2y^3)^{5 }}{(3x)^5} \\\\\\ =\frac{32y^{15}}{243x^5}$

5 0

2 years ago

Need help with math problems

mrs_skeptik [129]

Ok like what kind of math problems

3 0

2 years ago