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3 years ago

What is 12x + 1 x? ANSWER ASAPP PLZZZZ

Mathematics

2 answers:

Hoochie [10]3 years ago

8 0

Answer:

Hello!!! Princess Sakura here ^^

Step-by-step explanation:

$12x+1x\\13x$

Alex3 years ago

6 0

Answer:

13x

Step-by-step explanation:

12x+1x

Combine like term

13x

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What is the answer to yhis promble (4x+1)(2x+3)

Nikitich [7]

(4x+1)(2x+3)=8x*2+12x+2x+3
= 8x*2+14x+3

5 0

3 years ago

Read 2 more answers

Which changes in surface water movement are likely to occur when

Digiron [165]

Answer:

my best answer is D

Step-by-step explanation: plant keep the soil in place to decrease runoff and infiltration

8 0

3 years ago

Evaluate: <img src="https://tex.z-dn.net/?f=%5Csqrt%5B-3%5D%7B27%7D" id="TexFormula1" title="\sqrt[-3]{27}" alt="\sqrt[-3]{27}"

GenaCL600 [577]

Answer:

1/3 is the simplified form for given expression.

Step-by-step explanation:

Given that:

= $\sqrt[-3]{27}$

By simplifying:

Radical sign will be removed as follows:

= $27^{-1/3}$

For removing the "-" sign from power, base will be inverted:

= $(1/27) ^ {1/3}$

27 can also be written as 3 * 3 * 3 = 3^3

So,

= $(1/3^3)^{1/3}$

= $1^{1/3} / 3^{3* 1/3}$

By simplifying we get:

= 1/3

i hope it will help you!

6 0

3 years ago

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

3 years ago

A circle has an area of 78.5 square inches. What is the radius of the circle?

Rainbow [258]

Answer:

$\sqrt{24.98732606$ Decimal Form:

4.99873244

Step-by-step explanation:

geometry is the worst. but i have a way to remember how to find the area. (Apple Pies are too --> A=πr^2 --> area = pie×radius^squared) so basically you sub in 78.5 in for area and then you solve for r.

Hope i helped kinda lol

6 0

4 years ago