I will givee Brainlist if anyone answers all these questions

Mathematics

2 answers:

vaieri [72.5K]3 years ago

5 0

Theirs no picture, I dong know if you forgot to put it along with it but theirs nothing to answer

Troyanec [42]3 years ago

5 0

Answer:

There i no questions to answer on here

Step-by-step explanation:

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SURDS

aev [14]

1) √3 √7 = √21
2) √5 √245 = √5 √5 * 49 = √5 * 7√5 = 7 √5 * 5 = 7 √25 = 7 * 5 = 35
3) √77 ÷ √11 = as is. can't be simplified.
4) (√59)² = 59 ; the square root was cancelled by squared.
5) 3√6 x 8√7 = 3 * 8 √6 * 7 = 24 √42
6) 5√3 x 6 √3 = 5 * 6 √3 * 3 = 30 √9 = 30 * 3 = 90
7) 40√30 ÷ 5√3 = (40 / 5) * (√30 /√3) = 8 * ((√3 *10) / √3) = 8 √10
8) (6√5)² = 6² * √5² = 36 * 5 = 180

4 0

3 years ago

Which of these expressions are equivalent to 4x+2(2x−5)−(3−5x) ? Choose the TWO correct answers. 3x−13

Julli [10]

Answer: 4x+2(2x−5)−(3−5x) = 8x-10-3+5x

Step-by-step explanation: We need to find the equivalent expression to 4x+2(2x−5)−(3−5x).

Using distributive property as follows :

a(b+c) = ab+ac

4x+2(2x−5)−(3−5x) = 4x+4x-10-3+5x

= 8x-10-3+5x

Option (e)

Hence, the correct option is (e) "8x-10-3+5x"

3 0

3 years ago

HELP ME !!! PLEASE and thank you

PIT_PIT [208]

$\dfrac{MN}{ML}=\dfrac{HG}{HI}\\
\dfrac{MN}{18}=\dfrac{12}{27}\\
\dfrac{MN}{18}=\dfrac{4}{9}\\
9MN=72\\
MN=8$

6 0

3 years ago

1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t

natta225 [31]

The area of the surface given by $\vec r_1(u,v)$ is 1. In terms of a surface integral, we have

$1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

By multiplying each component in $\vec r_1$ by 5, we have

$\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}$

and the same goes for the derivative with respect to $v$ . Then the area of the surface given by $\vec r_2(u,v)$ is

$\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}$