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3 years ago

What makes a pattern a multiplication pattern?

Mathematics

2 answers:

Alecsey [184]3 years ago

3 0

Answer:

A multiplication pattern is created when you are multiplying the same number by the same number and go adding up!!!XD

Hope this helped!!!XD

Also can I get brainliest please?

SashulF [63]3 years ago

3 0

step by step :mutiplaction pattern is like this 2*2*2*8*2*2*2*8

just like a pattern

<em>hope this helps!</em>

You might be interested in

(6+23) x (32-25) + 72<br> I just need to see what the answer is to check my work

Vlad1618 [11]

Answer:

275

Step-by-step explanation:

6+23=29 and 32-25=7

29 x 7 = 203

203 + 72 = 275

The final answer is 275.

Hope this helps!

7 0

3 years ago

5 divided 630 in partial qoutients

KIM [24]

I think it is 0.007936

5 0

4 years ago

A coffee shop spends $408 for an order of 17 cases of paper cups. If the coffee shop adds 5 more cases of paper cups to the orde

-Dominant- [34]

Answer:

$528

Step-by-step explanation:

If cost is proportional to order quantity, the cost for 17+5 = 22 cases of cups will be 22/17 times the cost for 17 cases:

(22/17)($408) = $528

7 0

4 years ago

Assume y≠60 which expression is equivalent to (7sqrtx2)/(5sqrty3)

Drupady [299]

Answer:

The equivalent will be:

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$

Therefore, option 'a' is true.

Step-by-step explanation:

Given the expression

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}$

Let us solve the expression step by step to get the equivalent

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}$

$\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}$ ∵ $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=x^{2\cdot \frac{1}{7}}$

$=x^{\frac{2}{7}}$

also

$\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}$ ∵ $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=y^{3\cdot \frac{1}{5}}$

$=y^{\frac{3}{5}}$

so the expression becomes

$\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

$=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$ ∵ $\:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}$

Thus, the equivalent will be:

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$

Therefore, option 'a' is true.

5 0

3 years ago

If P(-9, -4), Q(-7, -1), R(-2, 5), S(-6, -1) are the coordinates of the points, what is the slope of line PQ?

matrenka [14]

Given :

P(-9, -4), Q(-7, -1), R(-2, 5), S(-6, -1).

To Find :

The slope of line PQ.

Solution :

We know , slope is given by :

$m=\dfrac{y_2-y_1}{x_2-x_1}\\\\m=\dfrac{-9-(-7)}{-4-(-1)}\\\\m=\dfrac{2}{3}$

Therefore, the slope of line PQ is 2/3.

Hence, this is the required solution.

8 0

3 years ago