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3 years ago

Please help! I will mark as brainliest. <3

Mathematics

1 answer:

DIA [1.3K]3 years ago

5 0

Answer:

185 greater than or equal to 53x

Step-by-step explanation:

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How can I show the work for 14.4 divided by 90? I forgot how to show it

kolezko [41]

The answer to this problem is 6.25 but in order to show the work you would do <span>90/14.4 and you would just have to work it out.

1.) You would have to move the decimal from 14.4 over one place to make it 144 and also the same for 90, making 900.
2.) Then you would just divide like normal.

</span>

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3 years ago

What is -14x+28+6x=-44 and show your work pls and thankyou.

Fed [463]

Answer:

x=2

Step-by-step explanation:

-14x+28+6x=-44

-8x+28=-44

-8x= -16

x=2

6 0

3 years ago

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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.

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3 years ago

Ubtract the sum of a2−2ab+b2 and 2a2+2ab+b2 from the sum of a2−b2 and a2+ab+3b2

frosja888 [35]

$\text{First sum:}\\\\(a^2-2ab+b^2)+(2a^2+2ab+b^2)=(a^2+2a^2)+(-2ab+2ab)+(b^2+b^2)=3a^2+2b^2\\\\\text{Second sum:}\\\\(a^2-b^2)+(a^2+ab+3b^2)=(a^2+a^2)+(ab)+(-b^2+3b^2)=2a^2+ab+2b^2\\\\\text{Difference:}\\\\(2a^2+ab+2b^2)-(3a^2+2b^2)=2a^2+ab+2b^2-3a^2-2b^2\\\\=(2a^2-3a^2)+(ab)+(2b^2-2b^2)=\boxed{-a^2+ab}$

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4 years ago

How would you describe the slice that created the triangular cross section on this cake

ahrayia [7]

perpendicular to the base

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3 years ago

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