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2 years ago

8

Decompose 35×10 to get an answer of 350

2 answers:

lapo4ka [179]2 years ago

8 0

I hope this helps you 35×10=5×7×2×5 25×14=350

nasty-shy [4]2 years ago

8 0

You already solve the questions

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A circle has a diameter of 12 inches. What is its area in terms of π?

Natalka [10]

Answer:

36π

Step-by-step explanation:

Since you have to use terms of pi you just leave out pi out of the formula

A = πr² so it would be

A = r² Radius is half of diameter so half of 12 = 6

A = 6²

A = 36

Than you add pi

36π

Hope this helped

4 0

3 years ago

If a/b = 4/3 and b/c = 2/3, what is a,b, and c?

kaheart [24]

Answer:

.

Step-by-step explanation:

$\frac{a}{b} = \frac{4}{3}$

$\frac{b}{c} = \frac{2}{3}$

to give b from both equation the same value

$\frac{a}{b} = \frac{4}{3} \times \frac{2}{2} = \frac{8}{6}$

$\frac{b}{c} = \frac{2}{3} \times \frac{3}{3} = \frac{6}{9}$

a:b:c = 8:6:9

a = 8

b = 6

c = 9

7 0

2 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

<em>C.</em> $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em> $(5-\frac{1}{2})^6$ <em />

3 0

2 years ago

Factor by grouping : xy + 6y - 2x - 12 (if you don't know the answer don't comment)

alexira [117]

<h3> Hey There today we will solve your problem</h3>

First we will factor out $y$ from $xy+6y$ which gives us $y(x+6)$

Next we will factor out $-2$ from $-2x-12$ which gives us $-2(x+6)$

This gives us the equation

$y(x+6)-2(x+6)$

then factor out the common term $x+6$ <em> and we get</em>

<em /> $\left(x+6\right)\left(y-2\right)$

6 0

2 years ago

How can I subtract 9 2/5 - 8 4/5

monitta

Answer:

0.6

Step-by-step explanation:

First make them into decimals so.

9.4 - 8.8 = 0.6

7 0

2 years ago

Read 2 more answers