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3 years ago

What number cannot be divided into 9 groups without a remainder?

Mathematics

1 answer:

Anna35 [415]3 years ago

4 0

the answer is b. 432

I did the first and got a reminder

I did the second and did not have a remainder so since that is the answer no need to do the rest!!

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Karina says every equilateral triangle is acute. Is this true explain

LenaWriter [7]

Yes, every equilateral triangle is acute because the angles are always congruent(identical) and 60 degrees which is acute. But I may be wrong. :P

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4 years ago

What is the percent of change from 64 to 48

Sladkaya [172]

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3 years ago

Please can I have an explanation also, I am terrible at these kinds of questions!

wlad13 [49]

Answer:

<em>The fraction of the beads that are red is</em>

Step-by-step explanation:

<u>Algebraic Expressions</u>

A bag contains red (r), yellow (y), and blue (b) beads. We are given the following ratios:

r:y = 2:3

y:b = 5:4

We are required to find r:s, where s is the total of beads in the bag, or

s = r + y + b

Thus, we need to calculate:

$\displaystyle \frac{r}{r+y+b} \qquad\qquad [1]$

Knowing that:

$\displaystyle \frac{r}{y}=\frac{2}{3} \qquad\qquad [2]$

$\displaystyle \frac{y}{b}=\frac{5}{4}$

Multiplying the equations above:

$\displaystyle \frac{r}{y}\frac{y}{b}=\frac{2}{3}\frac{5}{4}$

Simplifying:

$\displaystyle \frac{r}{b}=\frac{5}{6} \qquad\qquad [3]$

Dividing [1] by r:

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{1+y/r+b/r}$

Substituting from [2] and [3]:

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{1+3/2+6/5}$

Operating:

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{\frac{10+3*5+6*2}{10}}$

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{10}{10+15+12}$

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{10}{37}$

The fraction of the beads that are red is $\mathbf{\frac{10}{37}}$

8 0

3 years ago

Need the answer to the question

Morgarella [4.7K]

Answer:

The graph is wider and is reflected across the x axis.

Step-by-step explanation:

See attached image.

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2 years ago

Solve for W in A = 2(L + W)

Roman55 [17]

$\Large\begin{aligned}\\\\&A = 2(L + W)\\&A = 2L+2W\\&2W=A-2L\\&W=\dfrac{A-2L}{2}\end$

4 0

2 years ago